Use the formula $a \sin B \theta + b \cos B \theta = \sqrt { a ^ { 2 } + b ^ { 2 } } \sin ( B \theta + C )$ ,where $C = \arctan ( b / a ) , a = 18 , b = 6 , B = 3$ ,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) $6 \sqrt { 10 }$ $\sin ( \theta - 0.3218 )$
B) $6 \sqrt { 10 }$ $\sin ( 3 \theta + 0.3218 )$
C) $\sin ( 3 \theta + 0.3218 )$
D) $6 \sqrt { 10 }$ $\sin ( 3 \theta - 0.3218 )$
E) $6 \sqrt { 10 }$ $\sin ( \theta + 0.3218 )$

Question

Use the formula $a \sin B \theta + b \cos B \theta = \sqrt { a ^ { 2 } + b ^ { 2 } } \sin ( B \theta + C )$ ,where $C = \arctan ( b / a ) , a = 18 , b = 6 , B = 3$ ,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) $6 \sqrt { 10 }$ $\sin ( \theta - 0.3218 )$
B) $6 \sqrt { 10 }$ $\sin ( 3 \theta + 0.3218 )$
C) $\sin ( 3 \theta + 0.3218 )$
D) $6 \sqrt { 10 }$ $\sin ( 3 \theta - 0.3218 )$
E) $6 \sqrt { 10 }$ $\sin ( \theta + 0.3218 )$

Quizplus · Accepted Answer

Given $a = 18$, $b = 6$, and $B = 3$, we first calculate $\sqrt{a^2 + b^2} = \sqrt{18^2 + 6^2} = \sqrt{324 + 36} = \sqrt{360} = 6\sqrt{10}$. Then, $C = \arctan(b/a) = \arctan(6/18) = \arctan(1/3)$, which is approximately $0.3218$ radians. Therefore, the expression becomes $6\sqrt{10} \sin(3\theta + 0.3218)$.

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