Use the sum-to-product formulas to find the exact value of the given expression.
B)-1

Question

Use the sum-to-product formulas to find the exact value of the given expression. 
B)-1

Quizplus · Accepted Answer

Using the sum-to-product formula, we can rewrite the expression as:

$\sin(x)\cos(2x) + \cos(x)\sin(2x)$

$= \frac{1}{2}(\sin(x)\cos(2x) + \cos(x)\sin(2x)) + \frac{1}{2}(\sin(x)\cos(2x) - \cos(x)\sin(2x))$

$= \frac{1}{2}\sin(x + 2x) + \frac{1}{2}\sin(x - 2x)$

$= \frac{1}{2}\sin(3x) - \frac{1}{2}\sin(x)$

$= \sin\left(\frac{3x}{2}\right)\cos\left(\frac{x}{2}\right) - \cos\left(\frac{3x}{2}\right)\sin\left(\frac{x}{2}\right)$

$= -\sin\left(\frac{3x}{2} - \frac{\pi}{2}\right)\cos\left(\frac{x}{2}\right)$

$= -\sin\left(\frac{3}{2}(x - \pi)\right)\cos\left(\frac{x}{2}\right)$

$= -\sin\left(\frac{3}{2}(\frac{7\pi}{6})\right)\cos\left(\frac{\pi}{12}\right)$

$= -\sin\left(\frac{7\pi}{4}\right)\cos\left(\frac{\pi}{12}\right)$

$= -\left(-\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{6}+\sqrt{2}}{4}\right)$

$= \frac{\sqrt{6}+\sqrt{2}}{4}$

Therefore, the answer is C.

Use the Sum-To-Product Formulas to Find the Exact Value of the Given