# How to prove: cos^2(2x) - sin^2(x) = cos(x).cos(3x)?

Jun 26, 2018

impossible equation

#### Explanation:

${\cos}^{2} x - {\sin}^{2} x = \cos x . \cos 3 x$
Trig identity:
${\cos}^{2} x - {\sin}^{2} x = \cos 2 x$
$\cos 2 x = \cos x . \cos 3 x$.(1)
This equation is impossible.
If $x = \frac{\pi}{6}$ --> $2 x = \frac{\pi}{3}$ --> $\cos 2 x = \cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$ -->
$\cos 3 x = \cos \left(\frac{\pi}{2}\right) = 0$.
Equation (1) becomes:
1/2 = (sqrt3/2)(zero)
This is impossible.

Jun 26, 2018

First, develop the left side:
$f \left(x\right) = {\cos}^{2} \left(2 x\right) - {\sin}^{2} x = \left(\cos 2 x - \sin x\right) \left(\cos 2 x + \sin x\right)$
Use trig identities:
$\cos a + \cos b = 2 \cos \left(\frac{a + b}{2}\right) . \cos \left(\frac{a - b}{2}\right)$
$\cos a - \cos b = - 2 \sin \left(\frac{a + b}{2}\right) \sin \left(\frac{a - b}{2}\right)$
Note that:
cos 2x - sin x = cos 2x - cos (pi/2 - x) = -2sin(x/2 + pi/4)sin ((3x)/2 - pi/4).
(cos 2x + sin x) = cos 2x + cos (pi/2 - x) = = 2cos(x/2 + pi/4)cos ((3x)/2 - pi/4)
Also note that:
a. 2sin (x/2 + pi/4)(cos (x/2 + pi/4) = sin (x + pi/2) = cos x
b. $- 2 \cos \left(\frac{3 x}{2} - \frac{\pi}{4}\right) \sin \left(\frac{3 x}{2} - \frac{\pi}{4}\right) = - \sin \left(3 x - \frac{\pi}{2}\right) = \cos 3 x$
Finally:
$f \left(x\right) = {\cos}^{2} 2 x - {\sin}^{2} x = \cos x . \cos 3 x$. Proved