Deck 31: Using Fundamental Identities

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Question
Use the given values to evaluate (if possible)three trigonometric functions cos x,sin x,tan x.​ secx=4,sinx>0\sec x = 4 , \sin x > 0 ​ ​

A)​ cosx=14sinx=154tanx=15\begin{array} { l } \cos x = \frac { 1 } { 4 } \\\\\sin x = - \frac { \sqrt { 15 } } { 4 } \\\\\tan x = \sqrt { 15 }\end{array}
B)​ cosx=14sinx=154tanx=15\begin{array} { l } \cos x = \frac { 1 } { 4 } \\\\\sin x = \frac { \sqrt { 15 } } { 4 } \\\\\tan x = \sqrt { 15 }\end{array}
C)​ cosx=14sinx=154tanx=15\begin{array} { l } \cos x = \frac { 1 } { 4 } \\\sin x = - \frac { \sqrt { 15 } } { 4 } \\\tan x = - \sqrt { 15 }\end{array}
D)​ cosx=14sinx=154tanx=15\begin{array} { l } \cos x = - \frac { 1 } { 4 } \\\\\sin x = \frac { \sqrt { 15 } } { 4 } \\\\\tan x = \sqrt { 15 }\end{array}
E)​ cosx=14sinx=154tanx=15\begin{array} { l } \cos x = - \frac { 1 } { 4 } \\\sin x = - \frac { \sqrt { 15 } } { 4 } \\\tan x = - \sqrt { 15 }\end{array}
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Question
Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where 0<x<π20 < x < \frac { \pi } { 2 } .​ 16x2,x=4cosθ\sqrt { 16 - x ^ { 2 } } , x = 4 \cos \theta

A)16 cos θ
B)16 sin θ
C)4 sin θ
D)- 4 sin θ
E)4 cos θ
Question
Perform the multiplication and use the fundamental identities to simplify.​ (sinxcosx)2( \sin x - \cos x ) ^ { 2 }

A) 1+sinx+cosx1 + \sin x + \cos x
B) 22sinxcosx2 - 2 \sin x \cos x
C) 12sinxcosx1 - 2 \sin x \cos x
D) 1+2sinx+cosx1 + 2 \sin x + \cos x
E) 1sinxcosx1 - \sin x \cos x
Question
Use the given values to evaluate (if possible)three trigonometric functions cos x,csc x,tan x.​ sinx=15\sin x = \frac { 1 } { 5 } , cosx>0\cos x > 0

A)​ cosx=265cscx=5tanx=612\begin{array} { l } \cos x = - \frac { 2 \sqrt { 6 } } { 5 } \\\csc x = 5 \\\tan x = \frac { \sqrt { 6 } } { 12 }\end{array}
B)​ cosx=265cscx=5tanx=612\begin{array} { l } \cos x = \frac { 2 \sqrt { 6 } } { 5 } \\\csc x = 5 \\\tan x = - \frac { \sqrt { 6 } } { 12 }\end{array}
C)​ cosx=265cscx=5tanx=612\begin{array} { l } \cos x = - \frac { 2 \sqrt { 6 } } { 5 } \\\csc x = - 5 \\\tan x = - \frac { \sqrt { 6 } } { 12 }\end{array}
D)​ cosx=265cscx=5tanx=612\begin{array} { l } \cos x = \frac { 2 \sqrt { 6 } } { 5 } \\\csc x = 5 \\\tan x = \frac { \sqrt { 6 } } { 12 }\end{array}
E)​ cosx=265cscx=15tanx=612\begin{array} { l } \cos x = \frac { 2 \sqrt { 6 } } { 5 } \\\csc x = \frac { 1 } { 5 } \\\tan x = \frac { \sqrt { 6 } } { 12 }\end{array}
Question
Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where 0<θ<π20 < \theta < \frac { \pi } { 2 } . 369x2,x=2cosθ\sqrt { 36 - 9 x ^ { 2 } } , x = 2 \cos \theta

A)2 cos θ
B)36 sin θ
C)- 2 sin θ
D)36 cos θ
E)6 sin θ
Question
Use the given values to evaluate (if possible)three trigonometric functions csc x,tan x,cot x.​ cos(π2x)=817,cosx=1517\cos \left( \frac { \pi } { 2 } - x \right) = \frac { 8 } { 17 } , \cos x = \frac { 15 } { 17 }

A)​ cscx=85tanx=815cotx=158\begin{array} { l } \csc x = - \frac { 8 } { 5 } \\\\\tan x = - \frac { 8 } { 15 } \\\\\cot x = - \frac { 15 } { 8 }\end{array}
B)​ cscx=158tanx=815cotx=158\begin{array} { l } \csc x = - \frac { 15 } { 8 } \\\\\tan x = \frac { 8 } { 15 } \\\\\cot x = \frac { 15 } { 8 }\end{array}
C)​ cscx=158tanx=815cotx=158\begin{array} { l } \csc x = \frac { 15 } { 8 } \\\tan x = - \frac { 8 } { 15 } \\\cot x = \frac { 15 } { 8 }\end{array}
D)​ cscx=178tanx=815cotx=158\begin{array} { l } \csc x = \frac { 17 } { 8 } \\\\\tan x = \frac { 8 } { 15 } \\\\\cot x = \frac { 15 } { 8 }\end{array}
E)​ cscx=58tanx=815cotx=158\begin{array} { l } \csc x = \frac { 5 } { 8 } \\\\\tan x = \frac { 8 } { 15 } \\\\\cot x = - \frac { 15 } { 8 }\end{array}
Question
​Use the given values to evaluate (if possible)three trigonometric functions cscθ,tanθ,cosθ\csc \theta , \tan \theta , \cos \theta . ​​ sinθ=1,cotθ=0\sin \theta = - 1 , \cot \theta = 0

A)​ cscθ is undefined.tanθ=1cosθ=0\begin{array} { l } \csc \theta~ is~ undefined.\\\tan \theta = - 1 \\\cos \theta = 0\end{array}
B)​ cscθ=1tanθ is undefined.cosθ=1\csc \theta = - 1\\\tan \theta~ is~ undefined.\\\cos \theta = 1
C) cscθ=1tanθ is undefined.cosθ=0\csc \theta = - 1\\\tan \theta ~is~ undefined.\\\cos \theta = 0
D)​ cscθ=1tanθ is undefined.cosθ=1\csc \theta = - 1\\\tan \theta ~is~ undefined.\\\cos \theta = - 1
E)​ cscθ=1tanθ is undefined.cosθ=1\csc \theta = 1\\\tan \theta~ is~ undefined.\\\cos \theta = - 1
Question
Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where 0<θ<π20 < \theta < \frac { \pi } { 2 } . 16x2,x=4sinθ\sqrt { 16 - x ^ { 2 } } , x = 4 \sin \theta

A)4 sin θ
B)- 4 cos θ
C)4 cos θ
D)16 sin θ
E)16 cos θ
Question
Use the given values to evaluate (if possible)three trigonometric functions cos x,csc x,tan x.​ secx=3,sinx=63\sec x = \sqrt { 3 } , \sin x = - \frac { \sqrt { 6 } } { 3 }

A)​ cosx=33cscx=62\begin{array} { l } \cos x = \frac { \sqrt { 3 } } { 3 } \\\csc x = - \frac { \sqrt { 6 } } { 2 }\end{array} tan x = - 2\sqrt { 2 }

B)​ cosx=33cscx=62\begin{array} { l } \cos x = - \frac { \sqrt { 3 } } { 3 } \\\csc x = - \frac { \sqrt { 6 } } { 2 }\end{array} tan x = - ​ 2\sqrt { 2 }
C)​ cosx=66cscx=62\begin{array} { l } \cos x = \frac { \sqrt { 6 } } { 6 } \\\csc x = \frac { \sqrt { 6 } } { 2 }\end{array} ​tan x = 2\sqrt { 2 }

D)​ cosx=33cscx=62\begin{array} { l } \cos x = \frac { \sqrt { 3 } } { 3 } \\\csc x = - \frac { \sqrt { 6 } } { 2 }\end{array} ​tan x = 2\sqrt { 2 }

E)​ cosx=66cscx=62\begin{array} { l } \cos x = \frac { \sqrt { 6 } } { 6 } \\\csc x = \frac { \sqrt { 6 } } { 2 }\end{array} ​tan x = - 2\sqrt { 2 }
Question
Use the given values to evaluate (if possible)three trigonometric functions cotθ,secθ,cosθ\cot \theta , \sec \theta , \cos \theta .​ tanθ=4,cosθ<0\tan \theta = 4 , \cos \theta < 0

A)​ cotθ=14secθ=17cosθ=1717\begin{aligned}\cot \theta & = \frac { 1 } { 4 } \\\sec \theta & = - \sqrt { 17 } \\\cos \theta & = \frac { \sqrt { 17 } } { 17 }\end{aligned}
B)​ cotθ=4secθ=17cosθ=1717\begin{array} { l } \cot \theta = 4 \\\sec \theta = \sqrt { 17 } \\\cos \theta = \frac { \sqrt { 17 } } { 17 }\end{array}
C)​ cotθ=14secθ=17cosθ=1717\begin{array} { l } \cot \theta = \frac { 1 } { 4 } \\\sec \theta = - \sqrt { 17 } \\\cos \theta = - \frac { \sqrt { 17 } } { 17 }\end{array}
D)​ cotθ=14secθ=17cosθ=1717\begin{array} { l } \cot \theta = - \frac { 1 } { 4 } \\\sec \theta = - \sqrt { 17 } \\\cos \theta = - \frac { \sqrt { 17 } } { 17 }\end{array}
E)​ cotθ=14secθ=17cosθ=1717\begin{array} { l } \cot \theta = \frac { 1 } { 4 } \\\sec \theta = \sqrt { 17 } \\\cos \theta = - \frac { \sqrt { 17 } } { 17 }\end{array}
Question
By using a graphing utility to complete the following table.Round your answer to four decimal places. x0.20.40.60.811.21.4y1y2\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\\hline y _ { 2 } & \cdots & \cdots & \cdots & \ldots & \ldots & \cdots & \cdots \\\hline\end{array} y1=cos(π2x),y2=sinxy _ { 1 } = \cos \left( \frac { \pi } { 2 } - x \right) , y _ { 2 } = \sin x

A) x0.20.40.60.811.21.4y10.19870.38940.56460.71740.84150.93200.9854y20.19870.38940.56460.71740.84150.93200.9854\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & - 0.1987 & 0.3894 & 0.5646 & 0.7174 & 0.8415 & 0.9320 & 0.9854 \\\hline y _ { 2 } & - 0.1987 & 0.3894 & 0.5646 & 0.7174 & 0.8415 & 0.9320 & 0.9854 \\\hline\end{array}
B) x0.20.40.60.811.21.4y10.19870.38940.56460.71740.84150.93200.9854y20.19870.38940.56460.71740.84150.93200.9854\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & 0.1987 & - 0.3894 & 0.5646 & 0.7174 & 0.8415 & 0.9320 & 0.9854 \\\hline y _ { 2 } & 0.1987 & - 0.3894 & 0.5646 & 0.7174 & 0.8415 & 0.9320 & 0.9854 \\\hline\end{array}
C) x0.20.40.60.811.21.4y10.19870.38940.56460.71740.84150.93200.9854y20.19870.38940.56460.71740.84150.93200.9854\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & 0.1987 & 0.3894 & - 0.5646 & 0.7174 & 0.8415 & 0.9320 & 0.9854 \\\hline y _ { 2 } & 0.1987 & 0.3894 & - 0.5646 & 0.7174 & 0.8415 & 0.9320 & 0.9854 \\\hline\end{array}
D) x0.20.40.60.811.21.4y10.19870.38940.56460.71740.84150.93200.9854y20.19870.38940.56460.71740.84150.93200.9854\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & 0.1987 & 0.3894 & 0.5646 & - 0.7174 & 0.8415 & 0.9320 & 0.9854 \\\hline y _ { 2 } & 0.1987 & 0.3894 & 0.5646 & - 0.7174 & 0.8415 & 0.9320 & 0.9854 \\\hline\end{array}
E) x0.20.40.60.80.84150.93200.9854y10.19870.38940.56460.71740.84150.93200.9854y20.19870.38940.56460.71740.84150.93200.9854\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 0.8415 & 0.9320 & 0.9854 \\\hline y _ { 1 } & 0.1987 & 0.3894 & 0.5646 & 0.7174 & 0.8415 & 0.9320 & 0.9854 \\\hline y _ { 2 } & 0.1987 & 0.3894 & 0.5646 & 0.7174 & 0.8415 & 0.9320 & 0.9854 \\\hline\end{array}
Question
Use the given values to evaluate (if possible)three trigonometric functions sin x,cos x,cot x.​ cscx=419,tanx=940\csc x = \frac { 41 } { 9 } , \tan x = \frac { 9 } { 40 }

A)​ sinx=419cotx=409cosx=4041\begin{array} { l } \sin x = \frac { 41 } { 9 } \\\\\cot x = \frac { 40 } { 9 } \\\\\cos x = \frac { 40 } { 41 }\end{array}
B)​ sinx=409cotx=941cosx=4041\begin{array} { l } \sin x = \frac { 40 } { 9 } \\\\\cot x = \frac { 9 } { 41 } \\\\\cos x = \frac { 40 } { 41 }\end{array}
C)​ sinx=4041cotx=409cosx=941\begin{array} { l } \sin x = \frac { 40 } { 41 } \\\\\cot x = \frac { 40 } { 9 } \\\\\cos x = \frac { 9 } { 41 }\end{array}
D)​ sinx=941cotx=409cosx=4041\begin{array} { l } \sin x = \frac { 9 } { 41 } \\\\\cot x = \frac { 40 } { 9 } \\\\\cos x = \frac { 40 } { 41 }\end{array}
E)​ sinx=941cotx=4041cosx=409\begin{array} { l } \sin x = \frac { 9 } { 41 } \\\\\cot x = \frac { 40 } { 41 } \\\\\cos x = \frac { 40 } { 9 }\end{array}
Question
By using a graphing utility to complete the following table.Round your answer to four decimal places. x0.20.40.60.811.21.4y1y2\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & \cdots & \cdots & \ldots & \cdots & \cdots & \cdots & \cdots \\\hline y _ { 2 } & \cdots & \cdots & \ldots & \ldots & \cdots & \cdots & \cdots \\\hline\end{array} y1=secxcosx,y2=sinxtanxy _ { 1 } = \sec x - \cos x , y _ { 2 } = \sin x \tan x

A) x0.20.40.60.811.21.4y10.04030.16460.38630.73861.31052.39735.7135y20.04030.16460.38630.73861.31052.39735.7135\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & 0.0403 & 0.1646 & 0.3863 & 0.7386 & 1.3105 & 2.3973 & 5.7135 \\\hline y _ { 2 } & 0.0403 & 0.1646 & 0.3863 & 0.7386 & 1.3105 & 2.3973 & 5.7135 \\\hline\end{array}
B) x0.20.40.60.811.21.4y10.04030.16460.38630.73861.31052.39735.7135y20.04030.16460.38630.73861.31052.39735.7135\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & 0.0403 & 0.1646 & - 0.3863 & 0.7386 & 1.3105 & 2.3973 & 5.7135 \\\hline y _ { 2 } & 0.0403 & 0.1646 & - 0.3863 & 0.7386 & 1.3105 & 2.3973 & 5.7135 \\\hline\end{array}
C) x0.20.40.60.811.21.4y10.04030.16460.38630.73861.31052.39735.7135y20.04030.16460.38630.73861.31052.39735.7135\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & 0.0403 & - 0.1646 & 0.3863 & 0.7386 & 1.3105 & 2.3973 & 5.7135 \\\hline y _ { 2 } & 0.0403 & - 0.1646 & 0.3863 & 0.7386 & 1.3105 & 2.3973 & 5.7135 \\\hline\end{array}
D) x0.20.40.60.811.21.4y10.04030.16460.38630.73861.31052.39735.7135y20.04030.16460.38630.73861.31052.39735.7135\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & - 0.0403 & 0.1646 & 0.3863 & 0.7386 & 1.3105 & 2.3973 & 5.7135 \\\hline y _ { 2 } & - 0.0403 & 0.1646 & 0.3863 & 0.7386 & 1.3105 & 2.3973 & 5.7135 \\\hline\end{array}
E) x0.20.40.60.811.21.4y10.04030.16460.38630.73861.31052.39735.7135y20.04030.16460.38630.73861.31052.39735.7135\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & 0.0403 & 0.1646 & 0.3863 & - 0.7386 & 1.3105 & 2.3973 & 5.7135 \\\hline y _ { 2 } & 0.0403 & 0.1646 & 0.3863 & - 0.7386 & 1.3105 & 2.3973 & 5.7135 \\\hline\end{array}
Question
Use the given values to evaluate (if possible)three trigonometric functions cosϕ,sinϕ,tanϕ\cos \phi , \sin \phi , \tan \phi .​ secϕ=34,cscϕ=355\sec \phi = \frac { 3 } { 4 } , \csc \phi = \frac { - 3 \sqrt { 5 } } { 5 }

A)​ cosϕ=43sinϕ=53tanϕ=54\begin{array} { l } \cos \phi = \frac { 4 } { 3 } \\\sin \phi = - \frac { \sqrt { 5 } } { 3 } \\\tan \phi = \frac { \sqrt { 5 } } { 4 }\end{array}
B)​ cosϕ=43sinϕ=53tanϕ=54\begin{array} { l } \cos \phi = \frac { 4 } { 3 } \\\sin \phi = \frac { \sqrt { 5 } } { 3 } \\\tan \phi = - \frac { \sqrt { 5 } } { 4 }\end{array}
C)​ cosϕ=43sinϕ=53tanϕ=54\begin{array} { l } \cos \phi = \frac { 4 } { 3 } \\\sin \phi = - \frac { \sqrt { 5 } } { 3 } \\\tan \phi = - \frac { \sqrt { 5 } } { 4 }\end{array}
D)​ cosϕ=43sinϕ=53tanϕ=54\begin{array} { l } \cos \phi = \frac { 4 } { 3 } \\\sin \phi = \frac { \sqrt { 5 } } { 3 } \\\tan \phi = \frac { \sqrt { 5 } } { 4 }\end{array}
E)​ cosϕ=43sinϕ=53tanϕ=54\begin{array} { l } \cos \phi = - \frac { 4 } { 3 } \\\sin \phi = - \frac { \sqrt { 5 } } { 3 } \\\tan \phi = - \frac { \sqrt { 5 } } { 4 }\end{array}
Question
Use the given values to evaluate (if possible)two trigonometric functions tanφ\tan \varphi and cscψ\csc \psi .​ cotφ=5,sinψ=1010\cot \varphi = - 5 , \sin \psi = \frac { \sqrt { 10 } } { 10 }

A)​ tanφ=15cscψ=10\begin{array} { l } \tan \varphi = - \frac { 1 } { 5 } \\\csc \psi = \sqrt { 10 }\end{array}
B)​ tanφ=15cscψ=10\begin{array} { l } \tan \varphi = \frac { 1 } { 5 } \\\csc \psi = \sqrt { 10 }\end{array}
C)​ tanφ=15cscψ=10\begin{array} { l } \tan \varphi = \frac { 1 } { 5 } \\\csc \psi = - \sqrt { 10 }\end{array}
D)​ tanφ=5cscψ=10\begin{array} { l } \tan \varphi = - 5 \\\csc \psi = \sqrt { 10 }\end{array}
E)​ tanφ=15cscψ=10\begin{array} { l } \tan \varphi = - \frac { 1 } { 5 } \\\csc \psi = - \sqrt { 10 }\end{array}
Question
By using a graphing utility to complete the following table.Round your answer to two decimal places. x0.20.40.60.811.21.4y1y2\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & - & - & - & - & - & - & - \\\hline y _ { 2 } & - & - & - & - & - & - & - \\\hline\end{array} y1=sec4xsec2x,y2=tan2xsec2xy _ { 1 } = \sec ^ { 4 } x - \sec ^ { 2 } x , y _ { 2 } = \tan ^ { 2 } x \sec ^ { 2 } x

A) x0.20.40.60.811.21.4y10.040.210.692.188.3150.391,163.61y20.040.210.692.188.3150.391,163.61\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & - 0.04 & 0.21 & 0.69 & 2.18 & 8.31 & 50.39 & 1,163.61 \\\hline y _ { 2 } & - 0.04 & 0.21 & 0.69 & 2.18 & 8.31 & 50.39 & 1,163.61 \\\hline\end{array}
B) x0.20.40.60.811.21.4y10.040.210.692.188.3150.391,163.61y20.040.210.692.188.3150.391,163.61\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & 0.04 & 0.21 & 0.69 & 2.18 & 8.31 & 50.39 & 1,163.61 \\\hline y _ { 2 } & 0.04 & 0.21 & 0.69 & 2.18 & 8.31 & 50.39 & 1,163.61 \\\hline\end{array}
C) x0.20.40.60.811.21.4y10.040.210.692.188.3150.391,163.61y20.040.210.692.188.3150.391,163.61\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & 0.04 & 0.21 & 0.69 & - 2.18 & 8.31 & 50.39 & 1,163.61 \\\hline y _ { 2 } & 0.04 & 0.21 & 0.69 & - 2.18 & 8.31 & 50.39 & 1,163.61 \\\hline\end{array}
D) x0.20.40.60.811.21.4y10.040.210.692.188.3150.391,163.61y20.040.210.692.188.3150.391,163.61\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & 0.04 & 0.21 & - 0.69 & 2.18 & 8.31 & 50.39 & 1,163.61 \\\hline y _ { 2 } & 0.04 & 0.21 & - 0.69 & 2.18 & 8.31 & 50.39 & 1,163.61 \\\hline\end{array}
E) x0.20.40.60.811.21.4y10.040.210.692.188.3150.391,163.61y20.040.210.692.188.3150.391,163.61\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & 0.04 & 0.21 & 0.69 & 2.18 & 8.31 & 50.39 & - 1,163.61 \\\hline y _ { 2 } & 0.04 & 0.21 & 0.69 & 2.18 & 8.31 & 50.39 & - 1,163.61 \\\hline\end{array}
Question
Use the fundamental identities to simplify the expression.​ sinθtanθ\frac { \sin \theta } { \tan \theta }

A)cot θ
B)​cot θ
C)​sec θ
D)​cos θ
E)​csc θ
Question
By using a graphing utility to complete the following table.Round your answer to four decimal places. x0.81.01.21.41.61.82.0y1y2\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.8 & 1.0 & 1.2 & 1.4 & 1.6 & 1.8 & 2.0 \\\hline y _ { 1 } & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\\hline y _ { 2 } & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \cdots \\\hline\end{array} y1=cosx1sinx,y2=1+sinxcosxy _ { 1 } = \frac { \cos x } { 1 - \sin x } , y _ { 2 } = \frac { 1 + \sin x } { \cos x }

A) x0.81.01.21.41.61.82.0y12.46503.40825.331911.681468.47978.68764.5880y22.46503.40825.331911.681468.47978.68764.5880\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.8 & 1.0 & 1.2 & 1.4 & 1.6 & 1.8 & 2.0 \\\hline y _ { 1 } & 2.4650 & 3.4082 & 5.3319 & - 11.6814 & - 68.4797 & - 8.6876 & - 4.5880 \\\hline y _ { 2 } & 2.4650 & 3.4082 & 5.3319 & - 11.6814 & - 68.4797 & - 8.6876 & - 4.5880 \\\hline\end{array}
B) x0.81.01.21.41.61.82.0y12.46503.40825.331911.681468.47978.68764.5880y22.46503.40825.331911.681468.47978.68764.5880\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.8 & 1.0 & 1.2 & 1.4 & 1.6 & 1.8 & 2.0 \\\hline y _ { 1 } & - 2.4650 & 3.4082 & 5.3319 & 11.6814 & - 68.4797 & - 8.6876 & - 4.5880 \\\hline y _ { 2 } & - 2.4650 & 3.4082 & 5.3319 & 11.6814 & - 68.4797 & - 8.6876 & - 4.5880 \\\hline\end{array}
C) x0.81.01.21.41.61.82.0y2.46503.40825.331911.681468.47978.68764.5880y2.46503.40825.331911.681468.47978.68764.5880\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.8 & 1.0 & 1.2 & 1.4 & 1.6 & 1.8 & 2.0 \\\hline y & 2.4650 & 3.4082 & 5.3319 & 11.6814 & - - 68.4797 & - 8.6876 & - 4.5880 \\\hline y & 2.4650 & 3.4082 & 5.3319 & 11.6814 & - - 68.4797 & - 8.6876 & - 4.5880 \\\hline\end{array}
D) x0.81.01.21.41.61.82.0y12.46503.40825.331911.681468.47978.68764.5880y22.46503.40825.331911.681468.47978.68764.5880\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.8 & 1.0 & 1.2 & 1.4 & 1.6 & 1.8 & 2.0 \\\hline y _ { 1 } & 2.4650 & 3.4082 & - 5.3319 & 11.6814 & - 68.4797 & - 8.6876 & - 4.5880 \\\hline y _ { 2 } & 2.4650 & 3.4082 & - 5.3319 & 11.6814 & - 68.4797 & - 8.6876 & - 4.5880 \\\hline\end{array}
E) x0.81.01.21.41.61.82.0y12.46503.40825.331911.681468.47978.68764.5880y22.46503.40825.331911.681468.47978.68764.5880\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.8 & 1.0 & 1.2 & 1.4 & 1.6 & 1.8 & 2.0 \\\hline y _ { 1 } & 2.4650 & 3.4082 & 5.3319 & 11.6814 & - 68.4797 & - 8.6876 & - 4.5880 \\\hline y _ { 2 } & 2.4650 & 3.4082 & 5.3319 & 11.6814 & - 68.4797 & - 8.6876 & - 4.5880 \\\hline\end{array}
Question
Use the fundamental identities to simplify the expression. ​
Cot θ sec θ

A)tan θ
B)​sec θ cot θ
C)csc θ
D)cot θ
E)​cos θ
Question
Use the given values to evaluate (if possible)three trigonometric functions cos x,sin x,cot x.​ tanx=940,secx=4140\tan x = \frac { 9 } { 40 } , \sec x = - \frac { 41 } { 40 }

A) cosx=4041sinx=941cotx=409\begin{array} { l } \cos x = \frac { 40 } { 41 } \\\\\sin x = \frac { 9 } { 41 } \\\\\cot x = \frac { 40 } { 9 }\end{array}
B)​ cosx=4041sinx=941cotx=409\begin{array} { l } \cos x = - \frac { 40 } { 41 } \\\\\sin x = - \frac { 9 } { 41 } \\\\\cot x = - \frac { 40 } { 9 }\end{array}
C)​ cosx=4041sinx=941cotx=409\begin{array} { l } \cos x = - \frac { 40 } { 41 } \\\\\sin x = - \frac { 9 } { 41 } \\\\\cot x = \frac { 40 } { 9 }\end{array}
D)​ cosx=4041sinx=941cotx=409\begin{array} { l } \cos x = \frac { 40 } { 41 } \\\\\sin x = - \frac { 9 } { 41 } \\\\\cot x = - \frac { 40 } { 9 }\end{array}
E)​ cosx=4041sinx=941cotx=409\begin{array} { l } \cos x = - \frac { 40 } { 41 } \\\\\sin x = \frac { 9 } { 41 } \\\\\cot x = \frac { 40 } { 9 }\end{array}
Question
Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where π2<θ<π2- \frac { \pi } { 2 } < \theta < \frac { \pi } { 2 } .Then find sin θ and cos θ.​ 4=64x2,x=8sinθ4 = \sqrt { 64 - x ^ { 2 } } , x = 8 \sin \theta

A)​ 64sinθ=4;sinθ=±32;cosθ=1264 \sin \theta = 4 ; \sin \theta = \pm \frac { \sqrt { 3 } } { 2 } ; \cos \theta = \frac { 1 } { 2 }
B)​ 64cosθ=4;sinθ=±32;cosθ=1264 \cos \theta = 4 ; \sin \theta = \pm \frac { \sqrt { 3 } } { 2 } ; \cos \theta = \frac { 1 } { 2 }
C)​ 8sinθ=4;sinθ=±32;cosθ=128 \sin \theta = 4 ; \sin \theta = \pm \frac { \sqrt { 3 } } { 2 } ; \cos \theta = \frac { 1 } { 2 }
D)​ 8cosθ=4;sinθ=±32;cosθ=128 \cos \theta = 4 ; \sin \theta = \pm \frac { \sqrt { 3 } } { 2 } ; \cos \theta = \frac { 1 } { 2 }
E)​ 8cosθ=4;sinθ=12;cosθ=±328 \cos \theta = 4 ; \sin \theta = \frac { 1 } { 2 } ; \cos \theta = \pm \frac { \sqrt { 3 } } { 2 }
Question
Use fundamental identities to simplify the expression below and then determine which of the following is not equivalent. cosα(secαcosα)\cos \alpha ( \sec \alpha - \cos \alpha )

A)​ csc2αsec2α+tan2αcsc2α\frac { \csc ^ { 2 } \alpha - \sec ^ { 2 } \alpha + \tan ^ { 2 } \alpha } { \csc ^ { 2 } \alpha }
B)​ 1sin2α1 - \sin ^ { 2 } \alpha
C)​ csc2α1csc2α\frac { \csc ^ { 2 } \alpha - 1 } { \csc ^ { 2 } \alpha }
D)​ 1cos2α1 - \cos ^ { 2 } \alpha
E)​ 1cot2α1 - \cot ^ { 2 } \alpha
Question
Find the rate of change of the function f(x)=x+cosxf ( x ) = - x + \cos x . ​

A)​ sinx+1- \sin x + 1
B)​ sinx1\sin x - 1
C)​ sinx1- \sin x - 1
D)​ sinxx- \sin x - x
E)​ sinx+1\sin x + 1
Question
Find the rate of change of the function f(x)=cscxcosxf ( x ) = - \csc x - \cos x . ​

A)​ sinx+secx\sin x + \sec x
B)​ sinx+1\sin x + 1
C)​ sinxtanxsecx\sin x - \tan x \sec x
D)​ sinxsinx\sin x \sin x
E)​ cscxcotxsinx\csc x \cot x - \sin x
Question
Use the trigonometric substitution u = a csc θ,where 0 < θ < π/2 and a > 0 to simplify the expression u2a2\sqrt { u ^ { 2 } - a ^ { 2 } } . ​

A)a tan θ
B)- a cot θ
C)a csc θ
D)a cos θ
E)a cot θ
Question
Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where 0<x<π20 < x < \frac { \pi } { 2 } . 25x2+36,5x=6tanθ\sqrt { 25 x ^ { 2 } + 36 } , 5 x = 6 \tan \theta

A)25 tan θ
B)5 tan θ
C)25 sec θ
D)6 sec θ
E)- 5 sec θ
Question
Rewrite the expression as a single logarithm and simplify the result.​ lncosxlnsinx\ln | \cos x | - \ln | \sin x |

A)​ lntanx\ln | \tan x |
B) lncosx\ln | \cos x |
C) lnsinx\ln | \sin x |
D) lncotx\ln | \cot x |
E) lncscx\ln | \csc x |
Question
Use a calculator to demonstrate the identity for the value of θ.​ sin(θ)=sinθ,θ=258\sin ( - \theta ) = - \sin \theta , \theta = 258 ^ { \circ }

A)​ sin(258)=sin2580.9781\sin \left( - 258 ^ { \circ } \right) = - \sin 258 ^ { \circ } \approx 0.9781
B)​ sin(258)=sin2581.0081\sin \left( - 258 ^ { \circ } \right) = - \sin 258 ^ { \circ } \approx 1.0081
C)​ sin(258)=sin2580.9981\sin \left( - 258 ^ { \circ } \right) = - \sin 258 ^ { \circ } \approx 0.9981
D)​ sin(258)=sin2580.9781\sin \left( - 258 ^ { \circ } \right) = - \sin 258 ^ { \circ } \approx - 0.9781
E)​ sin(258)=sin2580.9981\sin \left( - 258 ^ { \circ } \right) = - \sin 258 ^ { \circ } \approx - 0.9981
Question
Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where π2<x<π2- \frac { \pi } { 2 } < x < \frac { \pi } { 2 } .Then find sin θ and cos θ. 4=16x2,x=4sinθ4 = \sqrt { 16 - x ^ { 2 } } , x = 4 \sin \theta

A) 16cosθ=4;sinθ=0;cosθ=116 \cos \theta = 4 ; \sin \theta = 0 ; \cos \theta = 1
B)​ 4cosθ=4;sinθ=0;cosθ=14 \cos \theta = 4 ; \sin \theta = 0 ; \cos \theta = 1
C)​ 16sinθ=4;sinθ=0;cosθ=116 \sin \theta = 4 ; \sin \theta = 0 ; \cos \theta = 1
D)​ 4cosθ=4;sinθ=1;cosθ=04 \cos \theta = 4 ; \sin \theta = 1 ; \cos \theta = 0
E)​ 4sinθ=4;sinθ=0;cosθ=14 \sin \theta = 4 ; \sin \theta = 0 ; \cos \theta = 1
Question
Find the rate of change of the function f(x)=secx+cscxf ( x ) = \sec x + \csc x . ​

A) cscxcotx+tanxsecx- \csc x \cot x + \tan x \sec x
B) cscxcotxcscxcotx\csc x \cot x - \csc x \cot x
C) cscxcotx+1- \csc x \cot x + 1
D) cscxcotxtanxsecx- \csc x \cot x - \tan x \sec x
E) cscxcotx+secx- \csc x \cot x + \sec x
Question
Use the trigonometric substitution u = a sin θ,where - π/2 < θ < π/2 and a > 0 to simplify the expression a2u2\sqrt { a ^ { 2 } - u ^ { 2 } } . ​

A)​a csc θ
B)​a sin θ
C)​a tan θ
D)​a cos θ
E)​- a cos θ
Question
Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where 0<θ<π20 < \theta < \frac { \pi } { 2 } . 11x2,x=11sinθ\sqrt { 11 - x ^ { 2 } } , x = \sqrt { 11 } \sin \theta

A) 11cosθ- \sqrt { 11 } \cos \theta
B) 11cosθ\sqrt { 11 } \cos \theta
C) 11sinθ\sqrt { 11 } \sin \theta
D) 11cosθ11 \cos \theta
E) 11sinθ11 \sin \theta
Question
Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where 0<x<π20 < x < \frac { \pi } { 2 } . x2+25,x=5tanθ\sqrt { x ^ { 2 } + 25 } , x = 5 \tan \theta

A)- 5 sec θ
B)5 tan θ
C)25 sec θ
D)25 tan θ
E)5 sec θ
Question
Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where 0<x<π20 < x < \frac { \pi } { 2 } . 6x2,x=6sinθ\sqrt { 6 - x ^ { 2 } } , x = \sqrt { 6 } \sin \theta

A)6 sin θ
B) 6cosθ- \sqrt { 6 } \cos \theta
C) 6sinθ\sqrt { 6 } \sin \theta
D) 6cosθ\sqrt { 6 } \cos \theta
E)6 cos θ
Question
Rewrite the expression as a single logarithm and simplify the result.​ lnsecx+lnsinx\ln | \sec x | + \ln | \sin x |

A) lncosx\ln | \cos x |
B) lncscx\ln | \csc x |
C) lncotx\ln | \cot x |
D) lnsinx\ln | \sin x |
E) lntanx\ln | \tan x |
Question
Use the trigonometric substitution u = a tan θ,where 0 < θ < π/2 and to simplify the expression a2+u2\sqrt { a ^ { 2 } + u ^ { 2 } } . ​

A)​- a sec θ
B)​a sec θ
C)​a tan θ
D)​a sin θ
E)​a csc θ
Question
Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where π2<θ<π2- \frac { \pi } { 2 } < \theta < \frac { \pi } { 2 } .Then find sin θ and cos θ.​ 33=819x2,x=3cosθ3 \sqrt { 3 } = \sqrt { 81 - 9 x ^ { 2 } } , x = 3 \cos \theta

A)​ sinθ=3;sinθ=33;cosθ=63\sin \theta = 3 ; \sin \theta = \frac { \sqrt { 3 } } { 3 } ; \cos \theta = \frac { \sqrt { 6 } } { 3 }
B)​ 3sinθ=3;sinθ=0;cosθ=13 \sin \theta = 3 ; \sin \theta = 0 ; \cos \theta = 1
C)​ sinθ=3;sinθ=33;cosθ=63\sin \theta = \sqrt { 3 } ; \sin \theta = \frac { \sqrt { 3 } } { 3 } ; \cos \theta = \frac { \sqrt { 6 } } { 3 }
D)​ 9sinθ=33;sinθ=33;cosθ=639 \sin \theta = 3 \sqrt { 3 } ; \sin \theta = \frac { \sqrt { 3 } } { 3 } ; \cos \theta = \frac { \sqrt { 6 } } { 3 }
E)​ 3cosθ=3;sinθ=1;cosθ=03 \cos \theta = 3 ; \sin \theta = 1 ; \cos \theta = 0
Question
If sinx=12 and cosx=32\sin x = \frac { 1 } { 2 } \text { and } \cos x = \frac { \sqrt { 3 } } { 2 } ,evaluate the following function. ​cot x

A)​cot x = 23\frac { 2 } { 3 } 3\sqrt { 3 }
B)​cot x = 13\frac { 1 } { 3 } 3\sqrt { 3 }
C)​cot x = 2
D)​cot x = 13\frac { 1 } { 3 }
E)​cot x = 3\sqrt { 3 }
Question
Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where 0<x<π20 < x < \frac { \pi } { 2 } . x216,x=4secθ\sqrt { x ^ { 2 } - 16 } , x = 4 \sec \theta

A)4 tan θ
B)16 sec θ
C)- 4 tan θ
D)4 sec θ
E)16 tan θ
Question
The forces acting on an object weighing W units on an inclined plane positioned at an angle of θ with the horizontal (see figure)are modeled by​ μWsinθ=Wtanθ\mu W \sin \theta = W \tan \theta ​ where μ is the coefficient of friction.Solve the equation for μ and simplify the result.​  <strong>The forces acting on an object weighing W units on an inclined plane positioned at an angle of θ with the horizontal (see figure)are modeled by​  \mu W \sin \theta = W \tan \theta  ​ where μ is the coefficient of friction.Solve the equation for μ and simplify the result.​   ​</strong> A)  \mu = \tan \theta  B)  \mu = \csc \theta  C)  \mu = \sin \theta  D)​  \mu = \sec \theta  E)  \mu = \cot \theta  <div style=padding-top: 35px>

A) μ=tanθ\mu = \tan \theta
B) μ=cscθ\mu = \csc \theta
C) μ=sinθ\mu = \sin \theta
D)​ μ=secθ\mu = \sec \theta
E) μ=cotθ\mu = \cot \theta
Question
Factor;then use fundamental identities to simplify the expression below and determine which of the following is not equivalent.​ cot2αcot2αcos2α\cot ^ { 2 } \alpha - \cot ^ { 2 } \alpha \cos ^ { 2 } \alpha

A)​ cos2α\cos ^ { 2 } \alpha
B)​ sin2(π2α)\sin ^ { 2 } \left( \frac { \pi } { 2 } - \alpha \right)
C)​ 1sin2α1 - \sin ^ { 2 } \alpha
D)​ cot2α\cot ^ { 2 } \alpha
E)​ 1sec2α\frac { 1 } { \sec ^ { 2 } \alpha }
Question
If x = 6 sin θ,use trigonometric substitution to write 36x2\sqrt { 36 - x ^ { 2 } } as a trigonometric function of θ,where 0<θ<π20 < \theta < \frac { \pi } { 2 } .

A)​6 csc θ
B)​6 sec θ
C)​6 sin θ
D)​6 cos θ
E)​6 tan θ
Question
Factor;then use fundamental identities to simplify the expression below and determine which of the following is not equivalent. ​​ cot2α+cot2αtan2α\cot ^ { 2 } \alpha + \cot ^ { 2 } \alpha \cdot \tan ^ { 2 } \alpha

A)​ csc2α\csc ^ { 2 } \alpha
B)​ 1+cot2α1 + \cot ^ { 2 } \alpha
C)​ sec2α\sec ^ { 2 } \alpha
D)​ 1/cos2(π2α)1 / \cos ^ { 2 } \left( \frac { \pi } { 2 } - \alpha \right)
E)​ 11cos2α\frac { 1 } { 1 - \cos ^ { 2 } \alpha }
Question
Rewrite lnsinθlncosθ\ln | \sin \theta | - \ln | \cos \theta | as a single logarithm and then simplify the result. ​

A)​ lncosθ\ln | \cos \theta |
B)​ lncscθ\ln | \csc \theta |
C)​ lnsecθ\ln | \sec \theta |
D)​ lntanθ\ln | \tan \theta |
E)​ lnsinθ\ln | \sin \theta |
Question
Which of the following is equivalent to the given expression? cos2x1+sinx\frac { \cos ^ { 2 } x } { 1 + \sin x }

A)​ tanx+cosx\tan x + \cos x
B)​ 1sinx1 - \sin x
C)​ cscx+cotx\csc x + \cot x
D)​ cotxsinx\cot x - \sin x
E)​ cotxcosx+tanx\cot x \cos x + \tan x
Question
Use fundamental identities to simplify the expression below and then determine which of the following is not equivalent.​ sinα(cscαsinα)\sin \alpha ( \csc \alpha - \sin \alpha )

A)​ 1tan2α1 - \tan ^ { 2 } \alpha
B)​ csc2α1csc2α\frac { \csc ^ { 2 } \alpha - 1 } { \csc ^ { 2 } \alpha }
C)​ csc2αsec2α+tan2αcsc2α\frac { \csc ^ { 2 } \alpha - \sec ^ { 2 } \alpha + \tan ^ { 2 } \alpha } { \csc ^ { 2 } \alpha }
D)​ cos2α\cos ^ { 2 } \alpha
Question
Multiply;then use fundamental identities to simplify the expression below and determine which of the following is not equivalent.​ (tanx+1)2( \tan x + 1 ) ^ { 2 }

A)​ tan2x+1\tan ^ { 2 } x + 1
B)​ sec2x+2tanx\sec ^ { 2 } x + 2 \tan x
C)​ 1+2sinxcosxcos2x\frac { 1 + 2 \sin x \cos x } { \cos ^ { 2 } x }
D)​ tan2x+2tanx+1\tan ^ { 2 } x + 2 \tan x + 1
E)​ sec2x(1+2sinxcosx)\sec ^ { 2 } x ( 1 + 2 \sin x \cos x )
Question
If x = 2 tan θ,use trigonometric substitution to write 4+x2\sqrt { 4 + x ^ { 2 } } as a trigonometric function of θ,where π2<θ<π2- \frac { \pi } { 2 } < \theta < \frac { \pi } { 2 } . ​

A)​2 csc θ
B)​2 sec θ
C)​2 cos θ
D)​2 tan θ
E)​2 sin θ
Question
If x = 4 cot θ,use trigonometric substitution to write 16+x2\sqrt { 16 + x ^ { 2 } } as a trigonometric function of θ,where 0 < θ < π.

A)​4 cos θ
B)​4 csc θ
C)​4 cot θ
D)​4 sec θ
E)​4 sin θ
Question
Use fundamental identities to simplify the expression below and then determine which of the following is not equivalent.​ sin(π2x)cscx\sin \left( \frac { \pi } { 2 } - x \right) \csc x

A)​-1
B)​ cosxsinx\frac { \cos x } { \sin x }
C)​cot x
D)​1/ tan x
E)​cos x csc x
Question
Use fundamental identities to simplify the expression below and then determine which of the following is not equivalent. sin(π2x)cscx\sin \left( \frac { \pi } { 2 } - x \right) \csc x

A)​ tanx\tan x
B)​ cosxcscx\cos \mathrm { x } \csc \mathrm { x }
C)​ cotx\cot x
D)​ cosxsinx\frac { \cos x } { \sin x }
E)​ 1tanx\frac { 1 } { \tan x }
Question
Multiply;then use fundamental identities to simplify the expression below and determine which of the following is not equivalent.​ (cotx+1)2( \cot x + 1 ) ^ { 2 }

A)​ cot2x+1\cot ^ { 2 } x + 1
B)​ cot2x+2cotx\cot ^ { 2 } x + 2 \cot x
C)​ cot2x+2cotx+1\cot ^ { 2 } x + 2 \cot x + 1
D)​ csc2x(1+2sinxcosx)\csc ^ { 2 } x ( 1 + 2 \sin x \cos x )
E)​ 1+2sinxcosxsin2x\frac { 1 + 2 \sin x \cos x } { \sin ^ { 2 } x }
Question
Factor;then use fundamental identities to simplify the expression below and determine which of the following is not equivalent. sin3xsin2xsinx+1\sin ^ { 3 } x - \sin ^ { 2 } x - \sin x + 1

A)​ 1+cos2xsin\frac { 1 + \cos ^ { 2 } x } { \sin }
B)​ cos2x(1sinx)\cos ^ { 2 } x ( 1 - \sin x )
C)​ cos4x1+sinx\frac { \cos ^ { 4 } x } { 1 + \sin x }
D)​ cos2xcos2xsinx\cos ^ { 2 } x - \cos ^ { 2 } x \sin x
E)​ 1sinxsec2x\frac { 1 - \sin x } { \sec ^ { 2 } x }
Question
Factor;then use fundamental identities to simplify the expression below and determine which of the following is not equivalent.​ sin3xsin2xsinx+1\sin ^ { 3 } x - \sin ^ { 2 } x - \sin x + 1

A)​ cos2xcos2xsinx\cos ^ { 2 } x - \cos ^ { 2 } x \sin x
B)​ 1sinxsec2x\frac { 1 - \sin x } { \sec ^ { 2 } x }
C)​ cos4x1+sinx\frac { \cos ^ { 4 } x } { 1 + \sin x }
D)​ cos2x(1sinx)\cos ^ { 2 } x ( 1 - \sin x )
E)​ 1+cos2xsinx\frac { 1 + \cos ^ { 2 } x } { \sin x }
Question
If x = 5 sin θ,use trigonometric substitution to write 25x2\sqrt { 25 - x ^ { 2 } } as a trigonometric function of θ,where π2<θ<π2- \frac { \pi } { 2 } < \theta < \frac { \pi } { 2 } .

A)​5 sin θ
B)​5 cos θ
C)​5 tan θ
D)​5 csc θ
E)​5 sec θ
Question
Which of the following is equivalent to the given expression? cos2x1+sinx\frac { \cos ^ { 2 } x } { 1 + \sin x }

A)​ cotxcosx+tanx\cot x \cos x + \tan x
B)​ cscx+cotx\csc x + \cot x
C)​ 1sinx1 - \sin x
D)​ tanx+cosx\tan x + \cos x
E)​ tanxcotxsinx\tan x \cot x - \sin x
Question
Use a graphing utility to determine which of the trigonometric functions is equal to the following expression. cscxsinxcotx\frac { \csc x - \sin x } { \cot x }

A)​ y = cos x
 <strong>Use a graphing utility to determine which of the trigonometric functions is equal to the following expression.  \frac { \csc x - \sin x } { \cot x } </strong> A)​ y = cos x   B)​ y = sin x 2 - 2π   2π  X _ { \text {scl } } = \frac { \pi } { 2 }  - 2 C)​ y = csc x   D)​ y = cot x 4 - 2π   2π  X _ { \text {scl } } = \frac { \pi } { 2 }  - 4 E)​ y = tan x   <div style=padding-top: 35px>
B)​ y = sin x
2
- 2π  <strong>Use a graphing utility to determine which of the trigonometric functions is equal to the following expression.  \frac { \csc x - \sin x } { \cot x } </strong> A)​ y = cos x   B)​ y = sin x 2 - 2π   2π  X _ { \text {scl } } = \frac { \pi } { 2 }  - 2 C)​ y = csc x   D)​ y = cot x 4 - 2π   2π  X _ { \text {scl } } = \frac { \pi } { 2 }  - 4 E)​ y = tan x   <div style=padding-top: 35px>
Xscl =π2X _ { \text {scl } } = \frac { \pi } { 2 }
- 2
C)​ y = csc x
 <strong>Use a graphing utility to determine which of the trigonometric functions is equal to the following expression.  \frac { \csc x - \sin x } { \cot x } </strong> A)​ y = cos x   B)​ y = sin x 2 - 2π   2π  X _ { \text {scl } } = \frac { \pi } { 2 }  - 2 C)​ y = csc x   D)​ y = cot x 4 - 2π   2π  X _ { \text {scl } } = \frac { \pi } { 2 }  - 4 E)​ y = tan x   <div style=padding-top: 35px>
D)​ y = cot x
4
- 2π  <strong>Use a graphing utility to determine which of the trigonometric functions is equal to the following expression.  \frac { \csc x - \sin x } { \cot x } </strong> A)​ y = cos x   B)​ y = sin x 2 - 2π   2π  X _ { \text {scl } } = \frac { \pi } { 2 }  - 2 C)​ y = csc x   D)​ y = cot x 4 - 2π   2π  X _ { \text {scl } } = \frac { \pi } { 2 }  - 4 E)​ y = tan x   <div style=padding-top: 35px>
Xscl =π2X _ { \text {scl } } = \frac { \pi } { 2 }
- 4
E)​ y = tan x
 <strong>Use a graphing utility to determine which of the trigonometric functions is equal to the following expression.  \frac { \csc x - \sin x } { \cot x } </strong> A)​ y = cos x   B)​ y = sin x 2 - 2π   2π  X _ { \text {scl } } = \frac { \pi } { 2 }  - 2 C)​ y = csc x   D)​ y = cot x 4 - 2π   2π  X _ { \text {scl } } = \frac { \pi } { 2 }  - 4 E)​ y = tan x   <div style=padding-top: 35px>
Question
If x = 3 cot θ,use trigonometric substitution to write 9+x2\sqrt { 9 + x ^ { 2 } } as a trigonometric function of θ,where 0<θ<π0 < \theta < \pi . ​

A)​3 sec θ
B)​3 csc θ
C)​3 cos θ
D)​3 sin θ
E)​3 cot θ
Question
Rewrite lnsecθ+lnsinθ\ln | \sec \theta | + \ln | \sin \theta | as a single logarithm and then simplify the result.

A)​ln |sin θ|
B)​ln |tan θ|
C)​ln |cos θ|
D)​ln |sec θ|
E)​ln |csc θ|
Question
If x = 2 tan θ,use trigonometric substitution to write 4+x2\sqrt { 4 + x ^ { 2 } } as a trigonometric function of θ,where 0<θ<π20 < \theta < \frac { \pi } { 2 } . ​

A)​2 cos θ
B)​2 tan θ
C)​2 sin θ
D)​2 sec θ
E)​2 csc θ
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Deck 31: Using Fundamental Identities
1
Use the given values to evaluate (if possible)three trigonometric functions cos x,sin x,tan x.​ secx=4,sinx>0\sec x = 4 , \sin x > 0 ​ ​

A)​ cosx=14sinx=154tanx=15\begin{array} { l } \cos x = \frac { 1 } { 4 } \\\\\sin x = - \frac { \sqrt { 15 } } { 4 } \\\\\tan x = \sqrt { 15 }\end{array}
B)​ cosx=14sinx=154tanx=15\begin{array} { l } \cos x = \frac { 1 } { 4 } \\\\\sin x = \frac { \sqrt { 15 } } { 4 } \\\\\tan x = \sqrt { 15 }\end{array}
C)​ cosx=14sinx=154tanx=15\begin{array} { l } \cos x = \frac { 1 } { 4 } \\\sin x = - \frac { \sqrt { 15 } } { 4 } \\\tan x = - \sqrt { 15 }\end{array}
D)​ cosx=14sinx=154tanx=15\begin{array} { l } \cos x = - \frac { 1 } { 4 } \\\\\sin x = \frac { \sqrt { 15 } } { 4 } \\\\\tan x = \sqrt { 15 }\end{array}
E)​ cosx=14sinx=154tanx=15\begin{array} { l } \cos x = - \frac { 1 } { 4 } \\\sin x = - \frac { \sqrt { 15 } } { 4 } \\\tan x = - \sqrt { 15 }\end{array}
cosx=14sinx=154tanx=15\begin{array} { l } \cos x = \frac { 1 } { 4 } \\\\\sin x = \frac { \sqrt { 15 } } { 4 } \\\\\tan x = \sqrt { 15 }\end{array}
2
Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where 0<x<π20 < x < \frac { \pi } { 2 } .​ 16x2,x=4cosθ\sqrt { 16 - x ^ { 2 } } , x = 4 \cos \theta

A)16 cos θ
B)16 sin θ
C)4 sin θ
D)- 4 sin θ
E)4 cos θ
4 sin θ
3
Perform the multiplication and use the fundamental identities to simplify.​ (sinxcosx)2( \sin x - \cos x ) ^ { 2 }

A) 1+sinx+cosx1 + \sin x + \cos x
B) 22sinxcosx2 - 2 \sin x \cos x
C) 12sinxcosx1 - 2 \sin x \cos x
D) 1+2sinx+cosx1 + 2 \sin x + \cos x
E) 1sinxcosx1 - \sin x \cos x
12sinxcosx1 - 2 \sin x \cos x
4
Use the given values to evaluate (if possible)three trigonometric functions cos x,csc x,tan x.​ sinx=15\sin x = \frac { 1 } { 5 } , cosx>0\cos x > 0

A)​ cosx=265cscx=5tanx=612\begin{array} { l } \cos x = - \frac { 2 \sqrt { 6 } } { 5 } \\\csc x = 5 \\\tan x = \frac { \sqrt { 6 } } { 12 }\end{array}
B)​ cosx=265cscx=5tanx=612\begin{array} { l } \cos x = \frac { 2 \sqrt { 6 } } { 5 } \\\csc x = 5 \\\tan x = - \frac { \sqrt { 6 } } { 12 }\end{array}
C)​ cosx=265cscx=5tanx=612\begin{array} { l } \cos x = - \frac { 2 \sqrt { 6 } } { 5 } \\\csc x = - 5 \\\tan x = - \frac { \sqrt { 6 } } { 12 }\end{array}
D)​ cosx=265cscx=5tanx=612\begin{array} { l } \cos x = \frac { 2 \sqrt { 6 } } { 5 } \\\csc x = 5 \\\tan x = \frac { \sqrt { 6 } } { 12 }\end{array}
E)​ cosx=265cscx=15tanx=612\begin{array} { l } \cos x = \frac { 2 \sqrt { 6 } } { 5 } \\\csc x = \frac { 1 } { 5 } \\\tan x = \frac { \sqrt { 6 } } { 12 }\end{array}
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5
Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where 0<θ<π20 < \theta < \frac { \pi } { 2 } . 369x2,x=2cosθ\sqrt { 36 - 9 x ^ { 2 } } , x = 2 \cos \theta

A)2 cos θ
B)36 sin θ
C)- 2 sin θ
D)36 cos θ
E)6 sin θ
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6
Use the given values to evaluate (if possible)three trigonometric functions csc x,tan x,cot x.​ cos(π2x)=817,cosx=1517\cos \left( \frac { \pi } { 2 } - x \right) = \frac { 8 } { 17 } , \cos x = \frac { 15 } { 17 }

A)​ cscx=85tanx=815cotx=158\begin{array} { l } \csc x = - \frac { 8 } { 5 } \\\\\tan x = - \frac { 8 } { 15 } \\\\\cot x = - \frac { 15 } { 8 }\end{array}
B)​ cscx=158tanx=815cotx=158\begin{array} { l } \csc x = - \frac { 15 } { 8 } \\\\\tan x = \frac { 8 } { 15 } \\\\\cot x = \frac { 15 } { 8 }\end{array}
C)​ cscx=158tanx=815cotx=158\begin{array} { l } \csc x = \frac { 15 } { 8 } \\\tan x = - \frac { 8 } { 15 } \\\cot x = \frac { 15 } { 8 }\end{array}
D)​ cscx=178tanx=815cotx=158\begin{array} { l } \csc x = \frac { 17 } { 8 } \\\\\tan x = \frac { 8 } { 15 } \\\\\cot x = \frac { 15 } { 8 }\end{array}
E)​ cscx=58tanx=815cotx=158\begin{array} { l } \csc x = \frac { 5 } { 8 } \\\\\tan x = \frac { 8 } { 15 } \\\\\cot x = - \frac { 15 } { 8 }\end{array}
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7
​Use the given values to evaluate (if possible)three trigonometric functions cscθ,tanθ,cosθ\csc \theta , \tan \theta , \cos \theta . ​​ sinθ=1,cotθ=0\sin \theta = - 1 , \cot \theta = 0

A)​ cscθ is undefined.tanθ=1cosθ=0\begin{array} { l } \csc \theta~ is~ undefined.\\\tan \theta = - 1 \\\cos \theta = 0\end{array}
B)​ cscθ=1tanθ is undefined.cosθ=1\csc \theta = - 1\\\tan \theta~ is~ undefined.\\\cos \theta = 1
C) cscθ=1tanθ is undefined.cosθ=0\csc \theta = - 1\\\tan \theta ~is~ undefined.\\\cos \theta = 0
D)​ cscθ=1tanθ is undefined.cosθ=1\csc \theta = - 1\\\tan \theta ~is~ undefined.\\\cos \theta = - 1
E)​ cscθ=1tanθ is undefined.cosθ=1\csc \theta = 1\\\tan \theta~ is~ undefined.\\\cos \theta = - 1
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8
Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where 0<θ<π20 < \theta < \frac { \pi } { 2 } . 16x2,x=4sinθ\sqrt { 16 - x ^ { 2 } } , x = 4 \sin \theta

A)4 sin θ
B)- 4 cos θ
C)4 cos θ
D)16 sin θ
E)16 cos θ
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9
Use the given values to evaluate (if possible)three trigonometric functions cos x,csc x,tan x.​ secx=3,sinx=63\sec x = \sqrt { 3 } , \sin x = - \frac { \sqrt { 6 } } { 3 }

A)​ cosx=33cscx=62\begin{array} { l } \cos x = \frac { \sqrt { 3 } } { 3 } \\\csc x = - \frac { \sqrt { 6 } } { 2 }\end{array} tan x = - 2\sqrt { 2 }

B)​ cosx=33cscx=62\begin{array} { l } \cos x = - \frac { \sqrt { 3 } } { 3 } \\\csc x = - \frac { \sqrt { 6 } } { 2 }\end{array} tan x = - ​ 2\sqrt { 2 }
C)​ cosx=66cscx=62\begin{array} { l } \cos x = \frac { \sqrt { 6 } } { 6 } \\\csc x = \frac { \sqrt { 6 } } { 2 }\end{array} ​tan x = 2\sqrt { 2 }

D)​ cosx=33cscx=62\begin{array} { l } \cos x = \frac { \sqrt { 3 } } { 3 } \\\csc x = - \frac { \sqrt { 6 } } { 2 }\end{array} ​tan x = 2\sqrt { 2 }

E)​ cosx=66cscx=62\begin{array} { l } \cos x = \frac { \sqrt { 6 } } { 6 } \\\csc x = \frac { \sqrt { 6 } } { 2 }\end{array} ​tan x = - 2\sqrt { 2 }
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10
Use the given values to evaluate (if possible)three trigonometric functions cotθ,secθ,cosθ\cot \theta , \sec \theta , \cos \theta .​ tanθ=4,cosθ<0\tan \theta = 4 , \cos \theta < 0

A)​ cotθ=14secθ=17cosθ=1717\begin{aligned}\cot \theta & = \frac { 1 } { 4 } \\\sec \theta & = - \sqrt { 17 } \\\cos \theta & = \frac { \sqrt { 17 } } { 17 }\end{aligned}
B)​ cotθ=4secθ=17cosθ=1717\begin{array} { l } \cot \theta = 4 \\\sec \theta = \sqrt { 17 } \\\cos \theta = \frac { \sqrt { 17 } } { 17 }\end{array}
C)​ cotθ=14secθ=17cosθ=1717\begin{array} { l } \cot \theta = \frac { 1 } { 4 } \\\sec \theta = - \sqrt { 17 } \\\cos \theta = - \frac { \sqrt { 17 } } { 17 }\end{array}
D)​ cotθ=14secθ=17cosθ=1717\begin{array} { l } \cot \theta = - \frac { 1 } { 4 } \\\sec \theta = - \sqrt { 17 } \\\cos \theta = - \frac { \sqrt { 17 } } { 17 }\end{array}
E)​ cotθ=14secθ=17cosθ=1717\begin{array} { l } \cot \theta = \frac { 1 } { 4 } \\\sec \theta = \sqrt { 17 } \\\cos \theta = - \frac { \sqrt { 17 } } { 17 }\end{array}
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11
By using a graphing utility to complete the following table.Round your answer to four decimal places. x0.20.40.60.811.21.4y1y2\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\\hline y _ { 2 } & \cdots & \cdots & \cdots & \ldots & \ldots & \cdots & \cdots \\\hline\end{array} y1=cos(π2x),y2=sinxy _ { 1 } = \cos \left( \frac { \pi } { 2 } - x \right) , y _ { 2 } = \sin x

A) x0.20.40.60.811.21.4y10.19870.38940.56460.71740.84150.93200.9854y20.19870.38940.56460.71740.84150.93200.9854\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & - 0.1987 & 0.3894 & 0.5646 & 0.7174 & 0.8415 & 0.9320 & 0.9854 \\\hline y _ { 2 } & - 0.1987 & 0.3894 & 0.5646 & 0.7174 & 0.8415 & 0.9320 & 0.9854 \\\hline\end{array}
B) x0.20.40.60.811.21.4y10.19870.38940.56460.71740.84150.93200.9854y20.19870.38940.56460.71740.84150.93200.9854\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & 0.1987 & - 0.3894 & 0.5646 & 0.7174 & 0.8415 & 0.9320 & 0.9854 \\\hline y _ { 2 } & 0.1987 & - 0.3894 & 0.5646 & 0.7174 & 0.8415 & 0.9320 & 0.9854 \\\hline\end{array}
C) x0.20.40.60.811.21.4y10.19870.38940.56460.71740.84150.93200.9854y20.19870.38940.56460.71740.84150.93200.9854\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & 0.1987 & 0.3894 & - 0.5646 & 0.7174 & 0.8415 & 0.9320 & 0.9854 \\\hline y _ { 2 } & 0.1987 & 0.3894 & - 0.5646 & 0.7174 & 0.8415 & 0.9320 & 0.9854 \\\hline\end{array}
D) x0.20.40.60.811.21.4y10.19870.38940.56460.71740.84150.93200.9854y20.19870.38940.56460.71740.84150.93200.9854\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & 0.1987 & 0.3894 & 0.5646 & - 0.7174 & 0.8415 & 0.9320 & 0.9854 \\\hline y _ { 2 } & 0.1987 & 0.3894 & 0.5646 & - 0.7174 & 0.8415 & 0.9320 & 0.9854 \\\hline\end{array}
E) x0.20.40.60.80.84150.93200.9854y10.19870.38940.56460.71740.84150.93200.9854y20.19870.38940.56460.71740.84150.93200.9854\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 0.8415 & 0.9320 & 0.9854 \\\hline y _ { 1 } & 0.1987 & 0.3894 & 0.5646 & 0.7174 & 0.8415 & 0.9320 & 0.9854 \\\hline y _ { 2 } & 0.1987 & 0.3894 & 0.5646 & 0.7174 & 0.8415 & 0.9320 & 0.9854 \\\hline\end{array}
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12
Use the given values to evaluate (if possible)three trigonometric functions sin x,cos x,cot x.​ cscx=419,tanx=940\csc x = \frac { 41 } { 9 } , \tan x = \frac { 9 } { 40 }

A)​ sinx=419cotx=409cosx=4041\begin{array} { l } \sin x = \frac { 41 } { 9 } \\\\\cot x = \frac { 40 } { 9 } \\\\\cos x = \frac { 40 } { 41 }\end{array}
B)​ sinx=409cotx=941cosx=4041\begin{array} { l } \sin x = \frac { 40 } { 9 } \\\\\cot x = \frac { 9 } { 41 } \\\\\cos x = \frac { 40 } { 41 }\end{array}
C)​ sinx=4041cotx=409cosx=941\begin{array} { l } \sin x = \frac { 40 } { 41 } \\\\\cot x = \frac { 40 } { 9 } \\\\\cos x = \frac { 9 } { 41 }\end{array}
D)​ sinx=941cotx=409cosx=4041\begin{array} { l } \sin x = \frac { 9 } { 41 } \\\\\cot x = \frac { 40 } { 9 } \\\\\cos x = \frac { 40 } { 41 }\end{array}
E)​ sinx=941cotx=4041cosx=409\begin{array} { l } \sin x = \frac { 9 } { 41 } \\\\\cot x = \frac { 40 } { 41 } \\\\\cos x = \frac { 40 } { 9 }\end{array}
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13
By using a graphing utility to complete the following table.Round your answer to four decimal places. x0.20.40.60.811.21.4y1y2\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & \cdots & \cdots & \ldots & \cdots & \cdots & \cdots & \cdots \\\hline y _ { 2 } & \cdots & \cdots & \ldots & \ldots & \cdots & \cdots & \cdots \\\hline\end{array} y1=secxcosx,y2=sinxtanxy _ { 1 } = \sec x - \cos x , y _ { 2 } = \sin x \tan x

A) x0.20.40.60.811.21.4y10.04030.16460.38630.73861.31052.39735.7135y20.04030.16460.38630.73861.31052.39735.7135\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & 0.0403 & 0.1646 & 0.3863 & 0.7386 & 1.3105 & 2.3973 & 5.7135 \\\hline y _ { 2 } & 0.0403 & 0.1646 & 0.3863 & 0.7386 & 1.3105 & 2.3973 & 5.7135 \\\hline\end{array}
B) x0.20.40.60.811.21.4y10.04030.16460.38630.73861.31052.39735.7135y20.04030.16460.38630.73861.31052.39735.7135\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & 0.0403 & 0.1646 & - 0.3863 & 0.7386 & 1.3105 & 2.3973 & 5.7135 \\\hline y _ { 2 } & 0.0403 & 0.1646 & - 0.3863 & 0.7386 & 1.3105 & 2.3973 & 5.7135 \\\hline\end{array}
C) x0.20.40.60.811.21.4y10.04030.16460.38630.73861.31052.39735.7135y20.04030.16460.38630.73861.31052.39735.7135\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & 0.0403 & - 0.1646 & 0.3863 & 0.7386 & 1.3105 & 2.3973 & 5.7135 \\\hline y _ { 2 } & 0.0403 & - 0.1646 & 0.3863 & 0.7386 & 1.3105 & 2.3973 & 5.7135 \\\hline\end{array}
D) x0.20.40.60.811.21.4y10.04030.16460.38630.73861.31052.39735.7135y20.04030.16460.38630.73861.31052.39735.7135\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & - 0.0403 & 0.1646 & 0.3863 & 0.7386 & 1.3105 & 2.3973 & 5.7135 \\\hline y _ { 2 } & - 0.0403 & 0.1646 & 0.3863 & 0.7386 & 1.3105 & 2.3973 & 5.7135 \\\hline\end{array}
E) x0.20.40.60.811.21.4y10.04030.16460.38630.73861.31052.39735.7135y20.04030.16460.38630.73861.31052.39735.7135\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & 0.0403 & 0.1646 & 0.3863 & - 0.7386 & 1.3105 & 2.3973 & 5.7135 \\\hline y _ { 2 } & 0.0403 & 0.1646 & 0.3863 & - 0.7386 & 1.3105 & 2.3973 & 5.7135 \\\hline\end{array}
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14
Use the given values to evaluate (if possible)three trigonometric functions cosϕ,sinϕ,tanϕ\cos \phi , \sin \phi , \tan \phi .​ secϕ=34,cscϕ=355\sec \phi = \frac { 3 } { 4 } , \csc \phi = \frac { - 3 \sqrt { 5 } } { 5 }

A)​ cosϕ=43sinϕ=53tanϕ=54\begin{array} { l } \cos \phi = \frac { 4 } { 3 } \\\sin \phi = - \frac { \sqrt { 5 } } { 3 } \\\tan \phi = \frac { \sqrt { 5 } } { 4 }\end{array}
B)​ cosϕ=43sinϕ=53tanϕ=54\begin{array} { l } \cos \phi = \frac { 4 } { 3 } \\\sin \phi = \frac { \sqrt { 5 } } { 3 } \\\tan \phi = - \frac { \sqrt { 5 } } { 4 }\end{array}
C)​ cosϕ=43sinϕ=53tanϕ=54\begin{array} { l } \cos \phi = \frac { 4 } { 3 } \\\sin \phi = - \frac { \sqrt { 5 } } { 3 } \\\tan \phi = - \frac { \sqrt { 5 } } { 4 }\end{array}
D)​ cosϕ=43sinϕ=53tanϕ=54\begin{array} { l } \cos \phi = \frac { 4 } { 3 } \\\sin \phi = \frac { \sqrt { 5 } } { 3 } \\\tan \phi = \frac { \sqrt { 5 } } { 4 }\end{array}
E)​ cosϕ=43sinϕ=53tanϕ=54\begin{array} { l } \cos \phi = - \frac { 4 } { 3 } \\\sin \phi = - \frac { \sqrt { 5 } } { 3 } \\\tan \phi = - \frac { \sqrt { 5 } } { 4 }\end{array}
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15
Use the given values to evaluate (if possible)two trigonometric functions tanφ\tan \varphi and cscψ\csc \psi .​ cotφ=5,sinψ=1010\cot \varphi = - 5 , \sin \psi = \frac { \sqrt { 10 } } { 10 }

A)​ tanφ=15cscψ=10\begin{array} { l } \tan \varphi = - \frac { 1 } { 5 } \\\csc \psi = \sqrt { 10 }\end{array}
B)​ tanφ=15cscψ=10\begin{array} { l } \tan \varphi = \frac { 1 } { 5 } \\\csc \psi = \sqrt { 10 }\end{array}
C)​ tanφ=15cscψ=10\begin{array} { l } \tan \varphi = \frac { 1 } { 5 } \\\csc \psi = - \sqrt { 10 }\end{array}
D)​ tanφ=5cscψ=10\begin{array} { l } \tan \varphi = - 5 \\\csc \psi = \sqrt { 10 }\end{array}
E)​ tanφ=15cscψ=10\begin{array} { l } \tan \varphi = - \frac { 1 } { 5 } \\\csc \psi = - \sqrt { 10 }\end{array}
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16
By using a graphing utility to complete the following table.Round your answer to two decimal places. x0.20.40.60.811.21.4y1y2\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & - & - & - & - & - & - & - \\\hline y _ { 2 } & - & - & - & - & - & - & - \\\hline\end{array} y1=sec4xsec2x,y2=tan2xsec2xy _ { 1 } = \sec ^ { 4 } x - \sec ^ { 2 } x , y _ { 2 } = \tan ^ { 2 } x \sec ^ { 2 } x

A) x0.20.40.60.811.21.4y10.040.210.692.188.3150.391,163.61y20.040.210.692.188.3150.391,163.61\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & - 0.04 & 0.21 & 0.69 & 2.18 & 8.31 & 50.39 & 1,163.61 \\\hline y _ { 2 } & - 0.04 & 0.21 & 0.69 & 2.18 & 8.31 & 50.39 & 1,163.61 \\\hline\end{array}
B) x0.20.40.60.811.21.4y10.040.210.692.188.3150.391,163.61y20.040.210.692.188.3150.391,163.61\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & 0.04 & 0.21 & 0.69 & 2.18 & 8.31 & 50.39 & 1,163.61 \\\hline y _ { 2 } & 0.04 & 0.21 & 0.69 & 2.18 & 8.31 & 50.39 & 1,163.61 \\\hline\end{array}
C) x0.20.40.60.811.21.4y10.040.210.692.188.3150.391,163.61y20.040.210.692.188.3150.391,163.61\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & 0.04 & 0.21 & 0.69 & - 2.18 & 8.31 & 50.39 & 1,163.61 \\\hline y _ { 2 } & 0.04 & 0.21 & 0.69 & - 2.18 & 8.31 & 50.39 & 1,163.61 \\\hline\end{array}
D) x0.20.40.60.811.21.4y10.040.210.692.188.3150.391,163.61y20.040.210.692.188.3150.391,163.61\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & 0.04 & 0.21 & - 0.69 & 2.18 & 8.31 & 50.39 & 1,163.61 \\\hline y _ { 2 } & 0.04 & 0.21 & - 0.69 & 2.18 & 8.31 & 50.39 & 1,163.61 \\\hline\end{array}
E) x0.20.40.60.811.21.4y10.040.210.692.188.3150.391,163.61y20.040.210.692.188.3150.391,163.61\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 & 1.4 \\\hline y _ { 1 } & 0.04 & 0.21 & 0.69 & 2.18 & 8.31 & 50.39 & - 1,163.61 \\\hline y _ { 2 } & 0.04 & 0.21 & 0.69 & 2.18 & 8.31 & 50.39 & - 1,163.61 \\\hline\end{array}
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17
Use the fundamental identities to simplify the expression.​ sinθtanθ\frac { \sin \theta } { \tan \theta }

A)cot θ
B)​cot θ
C)​sec θ
D)​cos θ
E)​csc θ
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18
By using a graphing utility to complete the following table.Round your answer to four decimal places. x0.81.01.21.41.61.82.0y1y2\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.8 & 1.0 & 1.2 & 1.4 & 1.6 & 1.8 & 2.0 \\\hline y _ { 1 } & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\\hline y _ { 2 } & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \cdots \\\hline\end{array} y1=cosx1sinx,y2=1+sinxcosxy _ { 1 } = \frac { \cos x } { 1 - \sin x } , y _ { 2 } = \frac { 1 + \sin x } { \cos x }

A) x0.81.01.21.41.61.82.0y12.46503.40825.331911.681468.47978.68764.5880y22.46503.40825.331911.681468.47978.68764.5880\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.8 & 1.0 & 1.2 & 1.4 & 1.6 & 1.8 & 2.0 \\\hline y _ { 1 } & 2.4650 & 3.4082 & 5.3319 & - 11.6814 & - 68.4797 & - 8.6876 & - 4.5880 \\\hline y _ { 2 } & 2.4650 & 3.4082 & 5.3319 & - 11.6814 & - 68.4797 & - 8.6876 & - 4.5880 \\\hline\end{array}
B) x0.81.01.21.41.61.82.0y12.46503.40825.331911.681468.47978.68764.5880y22.46503.40825.331911.681468.47978.68764.5880\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.8 & 1.0 & 1.2 & 1.4 & 1.6 & 1.8 & 2.0 \\\hline y _ { 1 } & - 2.4650 & 3.4082 & 5.3319 & 11.6814 & - 68.4797 & - 8.6876 & - 4.5880 \\\hline y _ { 2 } & - 2.4650 & 3.4082 & 5.3319 & 11.6814 & - 68.4797 & - 8.6876 & - 4.5880 \\\hline\end{array}
C) x0.81.01.21.41.61.82.0y2.46503.40825.331911.681468.47978.68764.5880y2.46503.40825.331911.681468.47978.68764.5880\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.8 & 1.0 & 1.2 & 1.4 & 1.6 & 1.8 & 2.0 \\\hline y & 2.4650 & 3.4082 & 5.3319 & 11.6814 & - - 68.4797 & - 8.6876 & - 4.5880 \\\hline y & 2.4650 & 3.4082 & 5.3319 & 11.6814 & - - 68.4797 & - 8.6876 & - 4.5880 \\\hline\end{array}
D) x0.81.01.21.41.61.82.0y12.46503.40825.331911.681468.47978.68764.5880y22.46503.40825.331911.681468.47978.68764.5880\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.8 & 1.0 & 1.2 & 1.4 & 1.6 & 1.8 & 2.0 \\\hline y _ { 1 } & 2.4650 & 3.4082 & - 5.3319 & 11.6814 & - 68.4797 & - 8.6876 & - 4.5880 \\\hline y _ { 2 } & 2.4650 & 3.4082 & - 5.3319 & 11.6814 & - 68.4797 & - 8.6876 & - 4.5880 \\\hline\end{array}
E) x0.81.01.21.41.61.82.0y12.46503.40825.331911.681468.47978.68764.5880y22.46503.40825.331911.681468.47978.68764.5880\begin{array} { | l | l | l | l | l | l | l | l | } \hline x & 0.8 & 1.0 & 1.2 & 1.4 & 1.6 & 1.8 & 2.0 \\\hline y _ { 1 } & 2.4650 & 3.4082 & 5.3319 & 11.6814 & - 68.4797 & - 8.6876 & - 4.5880 \\\hline y _ { 2 } & 2.4650 & 3.4082 & 5.3319 & 11.6814 & - 68.4797 & - 8.6876 & - 4.5880 \\\hline\end{array}
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19
Use the fundamental identities to simplify the expression. ​
Cot θ sec θ

A)tan θ
B)​sec θ cot θ
C)csc θ
D)cot θ
E)​cos θ
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20
Use the given values to evaluate (if possible)three trigonometric functions cos x,sin x,cot x.​ tanx=940,secx=4140\tan x = \frac { 9 } { 40 } , \sec x = - \frac { 41 } { 40 }

A) cosx=4041sinx=941cotx=409\begin{array} { l } \cos x = \frac { 40 } { 41 } \\\\\sin x = \frac { 9 } { 41 } \\\\\cot x = \frac { 40 } { 9 }\end{array}
B)​ cosx=4041sinx=941cotx=409\begin{array} { l } \cos x = - \frac { 40 } { 41 } \\\\\sin x = - \frac { 9 } { 41 } \\\\\cot x = - \frac { 40 } { 9 }\end{array}
C)​ cosx=4041sinx=941cotx=409\begin{array} { l } \cos x = - \frac { 40 } { 41 } \\\\\sin x = - \frac { 9 } { 41 } \\\\\cot x = \frac { 40 } { 9 }\end{array}
D)​ cosx=4041sinx=941cotx=409\begin{array} { l } \cos x = \frac { 40 } { 41 } \\\\\sin x = - \frac { 9 } { 41 } \\\\\cot x = - \frac { 40 } { 9 }\end{array}
E)​ cosx=4041sinx=941cotx=409\begin{array} { l } \cos x = - \frac { 40 } { 41 } \\\\\sin x = \frac { 9 } { 41 } \\\\\cot x = \frac { 40 } { 9 }\end{array}
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21
Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where π2<θ<π2- \frac { \pi } { 2 } < \theta < \frac { \pi } { 2 } .Then find sin θ and cos θ.​ 4=64x2,x=8sinθ4 = \sqrt { 64 - x ^ { 2 } } , x = 8 \sin \theta

A)​ 64sinθ=4;sinθ=±32;cosθ=1264 \sin \theta = 4 ; \sin \theta = \pm \frac { \sqrt { 3 } } { 2 } ; \cos \theta = \frac { 1 } { 2 }
B)​ 64cosθ=4;sinθ=±32;cosθ=1264 \cos \theta = 4 ; \sin \theta = \pm \frac { \sqrt { 3 } } { 2 } ; \cos \theta = \frac { 1 } { 2 }
C)​ 8sinθ=4;sinθ=±32;cosθ=128 \sin \theta = 4 ; \sin \theta = \pm \frac { \sqrt { 3 } } { 2 } ; \cos \theta = \frac { 1 } { 2 }
D)​ 8cosθ=4;sinθ=±32;cosθ=128 \cos \theta = 4 ; \sin \theta = \pm \frac { \sqrt { 3 } } { 2 } ; \cos \theta = \frac { 1 } { 2 }
E)​ 8cosθ=4;sinθ=12;cosθ=±328 \cos \theta = 4 ; \sin \theta = \frac { 1 } { 2 } ; \cos \theta = \pm \frac { \sqrt { 3 } } { 2 }
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22
Use fundamental identities to simplify the expression below and then determine which of the following is not equivalent. cosα(secαcosα)\cos \alpha ( \sec \alpha - \cos \alpha )

A)​ csc2αsec2α+tan2αcsc2α\frac { \csc ^ { 2 } \alpha - \sec ^ { 2 } \alpha + \tan ^ { 2 } \alpha } { \csc ^ { 2 } \alpha }
B)​ 1sin2α1 - \sin ^ { 2 } \alpha
C)​ csc2α1csc2α\frac { \csc ^ { 2 } \alpha - 1 } { \csc ^ { 2 } \alpha }
D)​ 1cos2α1 - \cos ^ { 2 } \alpha
E)​ 1cot2α1 - \cot ^ { 2 } \alpha
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23
Find the rate of change of the function f(x)=x+cosxf ( x ) = - x + \cos x . ​

A)​ sinx+1- \sin x + 1
B)​ sinx1\sin x - 1
C)​ sinx1- \sin x - 1
D)​ sinxx- \sin x - x
E)​ sinx+1\sin x + 1
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24
Find the rate of change of the function f(x)=cscxcosxf ( x ) = - \csc x - \cos x . ​

A)​ sinx+secx\sin x + \sec x
B)​ sinx+1\sin x + 1
C)​ sinxtanxsecx\sin x - \tan x \sec x
D)​ sinxsinx\sin x \sin x
E)​ cscxcotxsinx\csc x \cot x - \sin x
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25
Use the trigonometric substitution u = a csc θ,where 0 < θ < π/2 and a > 0 to simplify the expression u2a2\sqrt { u ^ { 2 } - a ^ { 2 } } . ​

A)a tan θ
B)- a cot θ
C)a csc θ
D)a cos θ
E)a cot θ
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26
Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where 0<x<π20 < x < \frac { \pi } { 2 } . 25x2+36,5x=6tanθ\sqrt { 25 x ^ { 2 } + 36 } , 5 x = 6 \tan \theta

A)25 tan θ
B)5 tan θ
C)25 sec θ
D)6 sec θ
E)- 5 sec θ
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27
Rewrite the expression as a single logarithm and simplify the result.​ lncosxlnsinx\ln | \cos x | - \ln | \sin x |

A)​ lntanx\ln | \tan x |
B) lncosx\ln | \cos x |
C) lnsinx\ln | \sin x |
D) lncotx\ln | \cot x |
E) lncscx\ln | \csc x |
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28
Use a calculator to demonstrate the identity for the value of θ.​ sin(θ)=sinθ,θ=258\sin ( - \theta ) = - \sin \theta , \theta = 258 ^ { \circ }

A)​ sin(258)=sin2580.9781\sin \left( - 258 ^ { \circ } \right) = - \sin 258 ^ { \circ } \approx 0.9781
B)​ sin(258)=sin2581.0081\sin \left( - 258 ^ { \circ } \right) = - \sin 258 ^ { \circ } \approx 1.0081
C)​ sin(258)=sin2580.9981\sin \left( - 258 ^ { \circ } \right) = - \sin 258 ^ { \circ } \approx 0.9981
D)​ sin(258)=sin2580.9781\sin \left( - 258 ^ { \circ } \right) = - \sin 258 ^ { \circ } \approx - 0.9781
E)​ sin(258)=sin2580.9981\sin \left( - 258 ^ { \circ } \right) = - \sin 258 ^ { \circ } \approx - 0.9981
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29
Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where π2<x<π2- \frac { \pi } { 2 } < x < \frac { \pi } { 2 } .Then find sin θ and cos θ. 4=16x2,x=4sinθ4 = \sqrt { 16 - x ^ { 2 } } , x = 4 \sin \theta

A) 16cosθ=4;sinθ=0;cosθ=116 \cos \theta = 4 ; \sin \theta = 0 ; \cos \theta = 1
B)​ 4cosθ=4;sinθ=0;cosθ=14 \cos \theta = 4 ; \sin \theta = 0 ; \cos \theta = 1
C)​ 16sinθ=4;sinθ=0;cosθ=116 \sin \theta = 4 ; \sin \theta = 0 ; \cos \theta = 1
D)​ 4cosθ=4;sinθ=1;cosθ=04 \cos \theta = 4 ; \sin \theta = 1 ; \cos \theta = 0
E)​ 4sinθ=4;sinθ=0;cosθ=14 \sin \theta = 4 ; \sin \theta = 0 ; \cos \theta = 1
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30
Find the rate of change of the function f(x)=secx+cscxf ( x ) = \sec x + \csc x . ​

A) cscxcotx+tanxsecx- \csc x \cot x + \tan x \sec x
B) cscxcotxcscxcotx\csc x \cot x - \csc x \cot x
C) cscxcotx+1- \csc x \cot x + 1
D) cscxcotxtanxsecx- \csc x \cot x - \tan x \sec x
E) cscxcotx+secx- \csc x \cot x + \sec x
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31
Use the trigonometric substitution u = a sin θ,where - π/2 < θ < π/2 and a > 0 to simplify the expression a2u2\sqrt { a ^ { 2 } - u ^ { 2 } } . ​

A)​a csc θ
B)​a sin θ
C)​a tan θ
D)​a cos θ
E)​- a cos θ
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32
Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where 0<θ<π20 < \theta < \frac { \pi } { 2 } . 11x2,x=11sinθ\sqrt { 11 - x ^ { 2 } } , x = \sqrt { 11 } \sin \theta

A) 11cosθ- \sqrt { 11 } \cos \theta
B) 11cosθ\sqrt { 11 } \cos \theta
C) 11sinθ\sqrt { 11 } \sin \theta
D) 11cosθ11 \cos \theta
E) 11sinθ11 \sin \theta
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33
Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where 0<x<π20 < x < \frac { \pi } { 2 } . x2+25,x=5tanθ\sqrt { x ^ { 2 } + 25 } , x = 5 \tan \theta

A)- 5 sec θ
B)5 tan θ
C)25 sec θ
D)25 tan θ
E)5 sec θ
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34
Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where 0<x<π20 < x < \frac { \pi } { 2 } . 6x2,x=6sinθ\sqrt { 6 - x ^ { 2 } } , x = \sqrt { 6 } \sin \theta

A)6 sin θ
B) 6cosθ- \sqrt { 6 } \cos \theta
C) 6sinθ\sqrt { 6 } \sin \theta
D) 6cosθ\sqrt { 6 } \cos \theta
E)6 cos θ
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35
Rewrite the expression as a single logarithm and simplify the result.​ lnsecx+lnsinx\ln | \sec x | + \ln | \sin x |

A) lncosx\ln | \cos x |
B) lncscx\ln | \csc x |
C) lncotx\ln | \cot x |
D) lnsinx\ln | \sin x |
E) lntanx\ln | \tan x |
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36
Use the trigonometric substitution u = a tan θ,where 0 < θ < π/2 and to simplify the expression a2+u2\sqrt { a ^ { 2 } + u ^ { 2 } } . ​

A)​- a sec θ
B)​a sec θ
C)​a tan θ
D)​a sin θ
E)​a csc θ
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37
Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where π2<θ<π2- \frac { \pi } { 2 } < \theta < \frac { \pi } { 2 } .Then find sin θ and cos θ.​ 33=819x2,x=3cosθ3 \sqrt { 3 } = \sqrt { 81 - 9 x ^ { 2 } } , x = 3 \cos \theta

A)​ sinθ=3;sinθ=33;cosθ=63\sin \theta = 3 ; \sin \theta = \frac { \sqrt { 3 } } { 3 } ; \cos \theta = \frac { \sqrt { 6 } } { 3 }
B)​ 3sinθ=3;sinθ=0;cosθ=13 \sin \theta = 3 ; \sin \theta = 0 ; \cos \theta = 1
C)​ sinθ=3;sinθ=33;cosθ=63\sin \theta = \sqrt { 3 } ; \sin \theta = \frac { \sqrt { 3 } } { 3 } ; \cos \theta = \frac { \sqrt { 6 } } { 3 }
D)​ 9sinθ=33;sinθ=33;cosθ=639 \sin \theta = 3 \sqrt { 3 } ; \sin \theta = \frac { \sqrt { 3 } } { 3 } ; \cos \theta = \frac { \sqrt { 6 } } { 3 }
E)​ 3cosθ=3;sinθ=1;cosθ=03 \cos \theta = 3 ; \sin \theta = 1 ; \cos \theta = 0
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38
If sinx=12 and cosx=32\sin x = \frac { 1 } { 2 } \text { and } \cos x = \frac { \sqrt { 3 } } { 2 } ,evaluate the following function. ​cot x

A)​cot x = 23\frac { 2 } { 3 } 3\sqrt { 3 }
B)​cot x = 13\frac { 1 } { 3 } 3\sqrt { 3 }
C)​cot x = 2
D)​cot x = 13\frac { 1 } { 3 }
E)​cot x = 3\sqrt { 3 }
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39
Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where 0<x<π20 < x < \frac { \pi } { 2 } . x216,x=4secθ\sqrt { x ^ { 2 } - 16 } , x = 4 \sec \theta

A)4 tan θ
B)16 sec θ
C)- 4 tan θ
D)4 sec θ
E)16 tan θ
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40
The forces acting on an object weighing W units on an inclined plane positioned at an angle of θ with the horizontal (see figure)are modeled by​ μWsinθ=Wtanθ\mu W \sin \theta = W \tan \theta ​ where μ is the coefficient of friction.Solve the equation for μ and simplify the result.​  <strong>The forces acting on an object weighing W units on an inclined plane positioned at an angle of θ with the horizontal (see figure)are modeled by​  \mu W \sin \theta = W \tan \theta  ​ where μ is the coefficient of friction.Solve the equation for μ and simplify the result.​   ​</strong> A)  \mu = \tan \theta  B)  \mu = \csc \theta  C)  \mu = \sin \theta  D)​  \mu = \sec \theta  E)  \mu = \cot \theta

A) μ=tanθ\mu = \tan \theta
B) μ=cscθ\mu = \csc \theta
C) μ=sinθ\mu = \sin \theta
D)​ μ=secθ\mu = \sec \theta
E) μ=cotθ\mu = \cot \theta
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41
Factor;then use fundamental identities to simplify the expression below and determine which of the following is not equivalent.​ cot2αcot2αcos2α\cot ^ { 2 } \alpha - \cot ^ { 2 } \alpha \cos ^ { 2 } \alpha

A)​ cos2α\cos ^ { 2 } \alpha
B)​ sin2(π2α)\sin ^ { 2 } \left( \frac { \pi } { 2 } - \alpha \right)
C)​ 1sin2α1 - \sin ^ { 2 } \alpha
D)​ cot2α\cot ^ { 2 } \alpha
E)​ 1sec2α\frac { 1 } { \sec ^ { 2 } \alpha }
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42
If x = 6 sin θ,use trigonometric substitution to write 36x2\sqrt { 36 - x ^ { 2 } } as a trigonometric function of θ,where 0<θ<π20 < \theta < \frac { \pi } { 2 } .

A)​6 csc θ
B)​6 sec θ
C)​6 sin θ
D)​6 cos θ
E)​6 tan θ
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43
Factor;then use fundamental identities to simplify the expression below and determine which of the following is not equivalent. ​​ cot2α+cot2αtan2α\cot ^ { 2 } \alpha + \cot ^ { 2 } \alpha \cdot \tan ^ { 2 } \alpha

A)​ csc2α\csc ^ { 2 } \alpha
B)​ 1+cot2α1 + \cot ^ { 2 } \alpha
C)​ sec2α\sec ^ { 2 } \alpha
D)​ 1/cos2(π2α)1 / \cos ^ { 2 } \left( \frac { \pi } { 2 } - \alpha \right)
E)​ 11cos2α\frac { 1 } { 1 - \cos ^ { 2 } \alpha }
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44
Rewrite lnsinθlncosθ\ln | \sin \theta | - \ln | \cos \theta | as a single logarithm and then simplify the result. ​

A)​ lncosθ\ln | \cos \theta |
B)​ lncscθ\ln | \csc \theta |
C)​ lnsecθ\ln | \sec \theta |
D)​ lntanθ\ln | \tan \theta |
E)​ lnsinθ\ln | \sin \theta |
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45
Which of the following is equivalent to the given expression? cos2x1+sinx\frac { \cos ^ { 2 } x } { 1 + \sin x }

A)​ tanx+cosx\tan x + \cos x
B)​ 1sinx1 - \sin x
C)​ cscx+cotx\csc x + \cot x
D)​ cotxsinx\cot x - \sin x
E)​ cotxcosx+tanx\cot x \cos x + \tan x
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46
Use fundamental identities to simplify the expression below and then determine which of the following is not equivalent.​ sinα(cscαsinα)\sin \alpha ( \csc \alpha - \sin \alpha )

A)​ 1tan2α1 - \tan ^ { 2 } \alpha
B)​ csc2α1csc2α\frac { \csc ^ { 2 } \alpha - 1 } { \csc ^ { 2 } \alpha }
C)​ csc2αsec2α+tan2αcsc2α\frac { \csc ^ { 2 } \alpha - \sec ^ { 2 } \alpha + \tan ^ { 2 } \alpha } { \csc ^ { 2 } \alpha }
D)​ cos2α\cos ^ { 2 } \alpha
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47
Multiply;then use fundamental identities to simplify the expression below and determine which of the following is not equivalent.​ (tanx+1)2( \tan x + 1 ) ^ { 2 }

A)​ tan2x+1\tan ^ { 2 } x + 1
B)​ sec2x+2tanx\sec ^ { 2 } x + 2 \tan x
C)​ 1+2sinxcosxcos2x\frac { 1 + 2 \sin x \cos x } { \cos ^ { 2 } x }
D)​ tan2x+2tanx+1\tan ^ { 2 } x + 2 \tan x + 1
E)​ sec2x(1+2sinxcosx)\sec ^ { 2 } x ( 1 + 2 \sin x \cos x )
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48
If x = 2 tan θ,use trigonometric substitution to write 4+x2\sqrt { 4 + x ^ { 2 } } as a trigonometric function of θ,where π2<θ<π2- \frac { \pi } { 2 } < \theta < \frac { \pi } { 2 } . ​

A)​2 csc θ
B)​2 sec θ
C)​2 cos θ
D)​2 tan θ
E)​2 sin θ
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49
If x = 4 cot θ,use trigonometric substitution to write 16+x2\sqrt { 16 + x ^ { 2 } } as a trigonometric function of θ,where 0 < θ < π.

A)​4 cos θ
B)​4 csc θ
C)​4 cot θ
D)​4 sec θ
E)​4 sin θ
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50
Use fundamental identities to simplify the expression below and then determine which of the following is not equivalent.​ sin(π2x)cscx\sin \left( \frac { \pi } { 2 } - x \right) \csc x

A)​-1
B)​ cosxsinx\frac { \cos x } { \sin x }
C)​cot x
D)​1/ tan x
E)​cos x csc x
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51
Use fundamental identities to simplify the expression below and then determine which of the following is not equivalent. sin(π2x)cscx\sin \left( \frac { \pi } { 2 } - x \right) \csc x

A)​ tanx\tan x
B)​ cosxcscx\cos \mathrm { x } \csc \mathrm { x }
C)​ cotx\cot x
D)​ cosxsinx\frac { \cos x } { \sin x }
E)​ 1tanx\frac { 1 } { \tan x }
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52
Multiply;then use fundamental identities to simplify the expression below and determine which of the following is not equivalent.​ (cotx+1)2( \cot x + 1 ) ^ { 2 }

A)​ cot2x+1\cot ^ { 2 } x + 1
B)​ cot2x+2cotx\cot ^ { 2 } x + 2 \cot x
C)​ cot2x+2cotx+1\cot ^ { 2 } x + 2 \cot x + 1
D)​ csc2x(1+2sinxcosx)\csc ^ { 2 } x ( 1 + 2 \sin x \cos x )
E)​ 1+2sinxcosxsin2x\frac { 1 + 2 \sin x \cos x } { \sin ^ { 2 } x }
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53
Factor;then use fundamental identities to simplify the expression below and determine which of the following is not equivalent. sin3xsin2xsinx+1\sin ^ { 3 } x - \sin ^ { 2 } x - \sin x + 1

A)​ 1+cos2xsin\frac { 1 + \cos ^ { 2 } x } { \sin }
B)​ cos2x(1sinx)\cos ^ { 2 } x ( 1 - \sin x )
C)​ cos4x1+sinx\frac { \cos ^ { 4 } x } { 1 + \sin x }
D)​ cos2xcos2xsinx\cos ^ { 2 } x - \cos ^ { 2 } x \sin x
E)​ 1sinxsec2x\frac { 1 - \sin x } { \sec ^ { 2 } x }
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54
Factor;then use fundamental identities to simplify the expression below and determine which of the following is not equivalent.​ sin3xsin2xsinx+1\sin ^ { 3 } x - \sin ^ { 2 } x - \sin x + 1

A)​ cos2xcos2xsinx\cos ^ { 2 } x - \cos ^ { 2 } x \sin x
B)​ 1sinxsec2x\frac { 1 - \sin x } { \sec ^ { 2 } x }
C)​ cos4x1+sinx\frac { \cos ^ { 4 } x } { 1 + \sin x }
D)​ cos2x(1sinx)\cos ^ { 2 } x ( 1 - \sin x )
E)​ 1+cos2xsinx\frac { 1 + \cos ^ { 2 } x } { \sin x }
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55
If x = 5 sin θ,use trigonometric substitution to write 25x2\sqrt { 25 - x ^ { 2 } } as a trigonometric function of θ,where π2<θ<π2- \frac { \pi } { 2 } < \theta < \frac { \pi } { 2 } .

A)​5 sin θ
B)​5 cos θ
C)​5 tan θ
D)​5 csc θ
E)​5 sec θ
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56
Which of the following is equivalent to the given expression? cos2x1+sinx\frac { \cos ^ { 2 } x } { 1 + \sin x }

A)​ cotxcosx+tanx\cot x \cos x + \tan x
B)​ cscx+cotx\csc x + \cot x
C)​ 1sinx1 - \sin x
D)​ tanx+cosx\tan x + \cos x
E)​ tanxcotxsinx\tan x \cot x - \sin x
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57
Use a graphing utility to determine which of the trigonometric functions is equal to the following expression. cscxsinxcotx\frac { \csc x - \sin x } { \cot x }

A)​ y = cos x
 <strong>Use a graphing utility to determine which of the trigonometric functions is equal to the following expression.  \frac { \csc x - \sin x } { \cot x } </strong> A)​ y = cos x   B)​ y = sin x 2 - 2π   2π  X _ { \text {scl } } = \frac { \pi } { 2 }  - 2 C)​ y = csc x   D)​ y = cot x 4 - 2π   2π  X _ { \text {scl } } = \frac { \pi } { 2 }  - 4 E)​ y = tan x
B)​ y = sin x
2
- 2π  <strong>Use a graphing utility to determine which of the trigonometric functions is equal to the following expression.  \frac { \csc x - \sin x } { \cot x } </strong> A)​ y = cos x   B)​ y = sin x 2 - 2π   2π  X _ { \text {scl } } = \frac { \pi } { 2 }  - 2 C)​ y = csc x   D)​ y = cot x 4 - 2π   2π  X _ { \text {scl } } = \frac { \pi } { 2 }  - 4 E)​ y = tan x
Xscl =π2X _ { \text {scl } } = \frac { \pi } { 2 }
- 2
C)​ y = csc x
 <strong>Use a graphing utility to determine which of the trigonometric functions is equal to the following expression.  \frac { \csc x - \sin x } { \cot x } </strong> A)​ y = cos x   B)​ y = sin x 2 - 2π   2π  X _ { \text {scl } } = \frac { \pi } { 2 }  - 2 C)​ y = csc x   D)​ y = cot x 4 - 2π   2π  X _ { \text {scl } } = \frac { \pi } { 2 }  - 4 E)​ y = tan x
D)​ y = cot x
4
- 2π  <strong>Use a graphing utility to determine which of the trigonometric functions is equal to the following expression.  \frac { \csc x - \sin x } { \cot x } </strong> A)​ y = cos x   B)​ y = sin x 2 - 2π   2π  X _ { \text {scl } } = \frac { \pi } { 2 }  - 2 C)​ y = csc x   D)​ y = cot x 4 - 2π   2π  X _ { \text {scl } } = \frac { \pi } { 2 }  - 4 E)​ y = tan x
Xscl =π2X _ { \text {scl } } = \frac { \pi } { 2 }
- 4
E)​ y = tan x
 <strong>Use a graphing utility to determine which of the trigonometric functions is equal to the following expression.  \frac { \csc x - \sin x } { \cot x } </strong> A)​ y = cos x   B)​ y = sin x 2 - 2π   2π  X _ { \text {scl } } = \frac { \pi } { 2 }  - 2 C)​ y = csc x   D)​ y = cot x 4 - 2π   2π  X _ { \text {scl } } = \frac { \pi } { 2 }  - 4 E)​ y = tan x
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58
If x = 3 cot θ,use trigonometric substitution to write 9+x2\sqrt { 9 + x ^ { 2 } } as a trigonometric function of θ,where 0<θ<π0 < \theta < \pi . ​

A)​3 sec θ
B)​3 csc θ
C)​3 cos θ
D)​3 sin θ
E)​3 cot θ
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59
Rewrite lnsecθ+lnsinθ\ln | \sec \theta | + \ln | \sin \theta | as a single logarithm and then simplify the result.

A)​ln |sin θ|
B)​ln |tan θ|
C)​ln |cos θ|
D)​ln |sec θ|
E)​ln |csc θ|
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60
If x = 2 tan θ,use trigonometric substitution to write 4+x2\sqrt { 4 + x ^ { 2 } } as a trigonometric function of θ,where 0<θ<π20 < \theta < \frac { \pi } { 2 } . ​

A)​2 cos θ
B)​2 tan θ
C)​2 sin θ
D)​2 sec θ
E)​2 csc θ
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