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 $a = 3$, $b = 4$, and $B = 1$, we calculate $\sqrt{a^2 + b^2} = \sqrt{9 + 16} = \sqrt{25} = 5$, and $C = \arctan(b/a) = \arctan(4/3)$. Since $\arctan(4/3)$ is approximately $0.3218$ radians, the expression becomes $\sqrt{10} \sin(\theta + 0.3218)$.

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