Use the formula $a \sin B \theta + b \cos B \theta = \sqrt { a ^ { 2 } + b ^ { 2 } } \cos ( B \theta - C )$ ,where $C = \arctan ( a / b ) , a 0$ ,to rewrite the trigonometric expression in the form. $\sqrt { 6 } \sin \left( \theta + \frac { \pi } { 4 } \right)$
A) $$\sqrt { 3 } \sin \theta - \sqrt { 3 } \cos \theta$$
B) $$\sqrt { 3 } \sin \theta + \cos \theta$$
C) $\sin \theta - \sqrt { 3 } \cos \theta$
D) $\sqrt { 3 } \sin \theta + \sqrt { 3 } \cos \theta$
E) $\sin \theta + \cos \theta$

Question

Use the formula $a \sin B \theta + b \cos B \theta = \sqrt { a ^ { 2 } + b ^ { 2 } } \cos ( B \theta - C )$ ,where $C = \arctan ( a / b ) , a > 0$ ,to rewrite the trigonometric expression in the form.​ $\sqrt { 6 } \sin \left( \theta + \frac { \pi } { 4 } \right)$ ​
A) $$\sqrt { 3 } \sin \theta - \sqrt { 3 } \cos \theta$$
B) $$\sqrt { 3 } \sin \theta + \cos \theta$$
C) $\sin \theta - \sqrt { 3 } \cos \theta$
D) $\sqrt { 3 } \sin \theta + \sqrt { 3 } \cos \theta$
E) $\sin \theta + \cos \theta$

Quizplus · Accepted Answer

Using the formula, we have:$\sqrt { 6 } \sin \left( \theta + \frac { \pi } { 4 } \right) = \sqrt { 6 } \left( \sin \theta \cos \frac { \pi } { 4 } + \cos \theta \sin \frac { \pi } { 4 } \right)$$\sqrt { 6 } \sin \left( \theta + \frac { \pi } { 4 } \right) = \sqrt { 6 } \left( \frac { \sqrt { 2 } } { 2 } \sin \theta + \frac { \sqrt { 2 } } { 2 } \cos \theta \right)$$\sqrt { 6 } \sin \left( \theta + \frac { \pi } { 4 } \right) = \sqrt { 3 } \sin \theta + \sqrt { 3 } \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 $a \sin B \theta + b \cos B \theta = \sqrt { a ^ { 2 } + b ^ { 2 } } \cos ( B \theta - C )$