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Use the Formula asinBθ+bcosBθ=a2+b2cos(BθC)a \sin B \theta + b \cos B \theta = \sqrt { a ^ { 2 } + b ^ { 2 } } \cos ( B \theta - C )

Question 36

Multiple Choice

Use the formula asinBθ+bcosBθ=a2+b2cos(BθC) 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 ) C=arctan(a/b) ,a>0C = \arctan ( a / b ) , a > 0 ,to rewrite the trigonometric expression in the form asinBθ+bcosBθa \sin B \theta + b \cos B \theta ​ 9 cos(θπ4) \cos \left( \theta - \frac { \pi } { 4 } \right)


A) 922cosθ\frac { 9 \sqrt { 2 } } { 2 } \cos \theta
B) 922sinθ922cosθ- \frac { 9 \sqrt { 2 } } { 2 } \sin \theta - \frac { 9 \sqrt { 2 } } { 2 } \cos \theta
C) 922sinθ+922cosθ- \frac { 9 \sqrt { 2 } } { 2 } \sin \theta + \frac { 9 \sqrt { 2 } } { 2 } \cos \theta
D) 922sinθ+922cosθ\frac { 9 \sqrt { 2 } } { 2 } \sin \theta + \frac { 9 \sqrt { 2 } } { 2 } \cos \theta
E) 922sinθ922cosθ\frac { 9 \sqrt { 2 } } { 2 } \sin \theta - \frac { 9 \sqrt { 2 } } { 2 } \cos \theta

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