Show that the equation is not an identity by finding a value of x for which both sides are defined but not equal.
-$\cos 2 x + \cos 6 x = 2 \cos 4 x \cos x$
A) $\frac { \pi } { 4 }$
B) 0
C) $2 \pi$
D) $- 2 \pi$

Question

Quizplus · Accepted Answer

For $x = \frac{\pi}{4}$, the left side becomes $\cos\left(2 \cdot \frac{\pi}{4}\right) + \cos\left(6 \cdot \frac{\pi}{4}\right) = \cos\left(\frac{\pi}{2}\right) + \cos\left(\frac{3\pi}{2}\right) = 0 + 0 = 0$. The right side becomes $2 \cos\left(4 \cdot \frac{\pi}{4}\right) \cos\left(\frac{\pi}{4}\right) = 2 \cos(\pi) \cos\left(\frac{\pi}{4}\right) = 2 \cdot (-1) \cdot \frac{\sqrt{2}}{2} = -\sqrt{2}$. Since the left side is 0 and the right side is $-\sqrt{2}$, they are not equal for $x = \frac{\pi}{4}$, showing the equation is not an identity.

Show That the Equation Is Not an Identity by Finding cos⁡2x+cos⁡6x=2cos⁡4xcos⁡x\cos 2 x + \cos 6 x = 2 \cos 4 x \cos xcos2x+cos6x=2cos4xcosx

Show That the Equation Is Not an Identity by Finding $\cos 2 x + \cos 6 x = 2 \cos 4 x \cos x$