Use the formula $\operatorname { asin } B \theta + b \cos B \theta = \sqrt { a ^ { 2 } + b ^ { 2 } } \sin ( B \theta + C )$ where $\bar { C } = \arctan ( b / a ) , a 0$ to find the trigonometric expression in the following forms.
$\begin{array} { l }
y = \sqrt { a ^ { 2 } + b ^ { 2 } } \sin ( B \theta + C ) \
2 \sin \theta + 7 \cos \theta
\end{array}$
A) $\sqrt { 53 } \sin ( \theta + 1.2925 )$
B) $\sqrt { 53 } \sin ( \theta - 1.2925 )$
C) $2 \sin ( \theta + 1.2925 )$
D) $7 \sin ( \theta + 1.2925 )$
E) $\sin ( \theta + 1.2925 )$

Question

Use the formula $\operatorname { asin } B \theta + b \cos B \theta = \sqrt { a ^ { 2 } + b ^ { 2 } } \sin ( B \theta + C )$ where $\bar { C } = \arctan ( b / a ) , a > 0$ to find the trigonometric expression in the following forms. 
$\begin{array} { l } 
y = \sqrt { a ^ { 2 } + b ^ { 2 } } \sin ( B \theta + C ) \
2 \sin \theta + 7 \cos \theta
\end{array}$ 
A) $\sqrt { 53 } \sin ( \theta + 1.2925 )$
B) $\sqrt { 53 } \sin ( \theta - 1.2925 )$
C) $2 \sin ( \theta + 1.2925 )$
D) $7 \sin ( \theta + 1.2925 )$
E) $\sin ( \theta + 1.2925 )$

Quizplus · Accepted Answer

Given $2 \sin \theta + 7 \cos \theta$, we identify $a = 2$ and $b = 7$. Thus, $\sqrt{a^2 + b^2} = \sqrt{4 + 49} = \sqrt{53}$. The angle $C = \arctan(b/a) = \arctan(7/2)$, which numerically approximates to $1.2925$ radians. Therefore, the expression is $\sqrt{53} \sin(\theta + 1.2925)$.

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