Use the formula $a \sin B \theta + b \cos B \theta = \sqrt { a ^ { 2 } + b ^ { 2 } } \cos ( B \theta - C )$ , where $C = \arctan ( a / b )$ $C = \arctan ( a / b ) , a 0$ , to rewrite the trigonometric expression in the form $a \sin B \theta + b \cos B \theta$ 9 $\cos \left( \theta - \frac { \pi } { 4 } \right)$
A) $\frac { 9 \sqrt { 2 } } { 2 } \cos \theta$
B) $- \frac { 9 \sqrt { 2 } } { 2 } \sin \theta - \frac { 9 \sqrt { 2 } } { 2 } \cos \theta$
C) $- \frac { 9 \sqrt { 2 } } { 2 } \sin \theta + \frac { 9 \sqrt { 2 } } { 2 } \cos \theta$
D) $\frac { 9 \sqrt { 2 } } { 2 } \sin \theta + \frac { 9 \sqrt { 2 } } { 2 } \cos \theta$
E) $\frac { 9 \sqrt { 2 } } { 2 } \sin \theta - \frac { 9 \sqrt { 2 } } { 2 } \cos \theta$

Question

Use the formula $a \sin B \theta + b \cos B \theta = \sqrt { a ^ { 2 } + b ^ { 2 } } \cos ( B \theta - C )$ , where $C = \arctan ( a / b )$ $C = \arctan ( a / b ) , a > 0$ , to rewrite the trigonometric expression in the form $a \sin B \theta + b \cos B \theta$   9 $\cos \left( \theta - \frac { \pi } { 4 } \right)$  
A) $\frac { 9 \sqrt { 2 } } { 2 } \cos \theta$
B) $- \frac { 9 \sqrt { 2 } } { 2 } \sin \theta - \frac { 9 \sqrt { 2 } } { 2 } \cos \theta$
C) $- \frac { 9 \sqrt { 2 } } { 2 } \sin \theta + \frac { 9 \sqrt { 2 } } { 2 } \cos \theta$
D) $\frac { 9 \sqrt { 2 } } { 2 } \sin \theta + \frac { 9 \sqrt { 2 } } { 2 } \cos \theta$
E) $\frac { 9 \sqrt { 2 } } { 2 } \sin \theta - \frac { 9 \sqrt { 2 } } { 2 } \cos \theta$

Quizplus · Accepted Answer

Given the expression is $9 \cos \left( \theta - \frac { \pi } { 4 } \right)$, we can use the cosine of a sum formula: $\cos(A - B) = \cos A \cos B + \sin A \sin B$. Here, $A = \theta$ and $B = \frac{\pi}{4}$, so $\cos B = \sin B = \frac{\sqrt{2}}{2}$. Thus, the expression becomes $9\left(\cos \theta \frac{\sqrt{2}}{2} + \sin \theta \frac{\sqrt{2}}{2}\right)$, which simplifies to $\frac { 9 \sqrt { 2 } } { 2 } \sin \theta + \frac { 9 \sqrt { 2 } } { 2 } \cos \theta$.

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