Use the formula $a \sin B \theta + b \cos B \theta = \sqrt { a ^ { 2 } + b ^ { 2 } } \cos ( B \theta - C )$ ,where $C = \arctan ( a / b ) , a = 3 , b = 7 , B = 2$ to rewrite the trigonometric expression in the following form. $y = \sqrt { a ^ { 2 } + b ^ { 2 } } \cos ( B \theta - C )$ $a \sin B \theta + b \cos B \theta$
A) $\sqrt { 58 }$ $\cos ( 2 \theta + 0.4049 )$
B) $\sqrt { 58 }$ $\cos ( 2 \theta - 0.4049 )$
C)7 $\cos ( 2 \theta - 0.4049 )$
D)3 $\cos ( 2 \theta - 0.4049 )$
E)3 $\cos ( 2 \theta + 0.4049 )$

Question

Use the formula $a \sin B \theta + b \cos B \theta = \sqrt { a ^ { 2 } + b ^ { 2 } } \cos ( B \theta - C )$ ,where $C = \arctan ( a / b ) , a = 3 , b = 7 , B = 2$ to rewrite the trigonometric expression in the following form.​ $y = \sqrt { a ^ { 2 } + b ^ { 2 } } \cos ( B \theta - C )$ ​ $a \sin B \theta + b \cos B \theta$ ​
A) $\sqrt { 58 }$ $\cos ( 2 \theta + 0.4049 )$
B) $\sqrt { 58 }$ $\cos ( 2 \theta - 0.4049 )$
C)7 $\cos ( 2 \theta - 0.4049 )$
D)3 $\cos ( 2 \theta - 0.4049 )$
E)3 $\cos ( 2 \theta + 0.4049 )$

Quizplus · Accepted Answer

Given $a = 3$, $b = 7$, and $B = 2$, we first find $\sqrt{a^2 + b^2} = \sqrt{3^2 + 7^2} = \sqrt{9 + 49} = \sqrt{58}$. Then, $C = \arctan(a/b) = \arctan(3/7)$. The correct form is $\sqrt{58} \cos(2\theta - C)$, where $C$ is positive, indicating a phase shift to the right, thus $2\theta - C$ is the correct argument form.

Use the Formula asin⁡Bθ+bcos⁡Bθ=a2+b2cos⁡(Bθ−C)a \sin B \theta + b \cos B \theta = \sqrt { a ^ { 2 } + b ^ { 2 } } \cos ( B \theta - C )asinBθ+bcosBθ=a2+b2​cos(Bθ−C)

Use the Formula $a \sin B \theta + b \cos B \theta = \sqrt { a ^ { 2 } + b ^ { 2 } } \cos ( B \theta - C )$