Complete the identity.
-$$\frac { \cos 4 x - \cos 8 x } { \cos 4 x + \cos 8 x } = \text { ? }$$
A) $\tan 2 x \tan 6 x$
B) $- \tan 6 x$
C) $\cot 6 x$
D) 0

Question

Quizplus · Accepted Answer

The identity can be simplified using the sum-to-product formulas. The formula for the difference and sum of cosines is: $$\cos A - \cos B = -2 \sin \left(\frac{A + B}{2}\right) \sin \left(\frac{A - B}{2}\right)$$ and $$\cos A + \cos B = 2 \cos \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)$$. Applying these to the given expression, we get: $$\frac{-2 \sin(6x) \sin(-2x)}{2 \cos(6x) \cos(-2x)} = \frac{-\sin(6x) \sin(2x)}{\cos(6x) \cos(2x)} = \tan(2x) \tan(6x)$$.

Complete the Identity cos⁡4x−cos⁡8xcos⁡4x+cos⁡8x= ? \frac { \cos 4 x - \cos 8 x } { \cos 4 x + \cos 8 x } = \text { ? }cos4x+cos8xcos4x−cos8x​= ?

Complete the Identity $\frac { \cos 4 x - \cos 8 x } { \cos 4 x + \cos 8 x } = \text { ? }$