Use the formula $a \sin B \theta + b \cos B \theta = \sqrt { a ^ { 2 } + b ^ { 2 } } \sin ( B \theta + C )$ , where $C = \arctan ( b / a ) , a = 3 , b = , B = 1$ to rewrite the trigonometric expression in the following form.
$y = \sqrt { a ^ { 2 } + b ^ { 2 } } \sin ( B \theta + C )$ $a \sin B \theta + b \cos B \theta$
A) $\sin ( \theta + 0.3218 )$
B) $\sin ( \theta - 0.3218 )$
C)3 $\sin ( \theta + 0.3218 )$
D) $\sqrt { 10 }$ $\sin ( \theta - 0.3218 )$
E) $\sqrt { 10 }$ $\sin ( \theta + 0.3218 )$

Question

Use the formula $a \sin B \theta + b \cos B \theta = \sqrt { a ^ { 2 } + b ^ { 2 } } \sin ( B \theta + C )$ , where $C = \arctan ( b / a ) , a = 3 , b = , B = 1$ to rewrite the trigonometric expression in the following form. 
 $y = \sqrt { a ^ { 2 } + b ^ { 2 } } \sin ( B \theta + C )$   $a \sin B \theta + b \cos B \theta$  
A) $\sin ( \theta + 0.3218 )$
B) $\sin ( \theta - 0.3218 )$
C)3 $\sin ( \theta + 0.3218 )$
D) $\sqrt { 10 }$ $\sin ( \theta - 0.3218 )$
E) $\sqrt { 10 }$ $\sin ( \theta + 0.3218 )$

Quizplus · Accepted Answer

Given $a = 3$ and $b = 4$, we find $\sqrt{a^2 + b^2} = \sqrt{9 + 16} = \sqrt{25} = 5$. The angle $C = \arctan(b/a) = \arctan(4/3)$, which is approximately $0.3218$ radians. Thus, the expression becomes $5\sin(\theta + 0.3218)$.

Use the Formula asin⁡Bθ+bcos⁡Bθ=a2+b2sin⁡(Bθ+C)a \sin B \theta + b \cos B \theta = \sqrt { a ^ { 2 } + b ^ { 2 } } \sin ( B \theta + C )asinBθ+bcosBθ=a2+b2​sin(Bθ+C)

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