Question

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

Answer 1

impossible equation

Explanation:

`cos^2 x - sin^2 x = cos x.cos 3x`
Trig identity:
`cos^2 x - sin^2 x = cos 2x`
`cos 2x = cos x.cos 3x`.(1)
This equation is impossible.
If `x = pi/6` --> `2x = pi/3` --> `cos 2x = cos (pi/3) = 1/2` -->
`cos 3x = cos (pi/2) = 0`.
Equation (1) becomes:
1/2 = (sqrt3/2)(zero)
This is impossible.

Answer 2

First, develop the left side:
`f(x) = cos^2 (2x) - sin^2 x = (cos 2x - sin x)(cos 2x + sin x)`
Use trig identities:
`cos a + cos b = 2cos ((a + b)/2).cos ((a - b)/2)`
`cos a - cos b = - 2sin ((a +b)/2)sin ((a - b)/2)`
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. `-2cos((3x)/2 - pi/4)sin ((3x)/2 - pi/4) = -sin (3x - pi/2) = cos 3x`
Finally:
`f(x) = cos^2 2x - sin^2 x = cos x.cos 3x`. Proved