Question

Question #95289

Answer 1

Use trig identities:
`cos a + cos b = 2cos ((a + b)/2)cos ((a - b)/2)`(1)
`cos^3 a = cos x(4cos^2 x - 3)` (2)
`sin 2x = 2sin x.cos x` (3)
`sin^2 x + cos^2 x = 1` (4)
In this case, from identity (1), we get:
f(x) = cos (4x) + cos (2x) = 2cos (3x).cos x
Substitute cos 3x by identity (2)
`f(x) = (2cos^2 x)(4cos^2 x - 3)`
Replace `cos^2 x` by `(1 - sin^2 x)`
`f(x) = 2cos^2 x( 4 - 4sin^2 x - 3) = 2cos^2 x(-4sin^2 x + 1)`
From identities (3) and (4), we get:
`f(x) = - 8sin^2 x.cos^2 x + 2cos^2 x = - 2sin^2 (2x) + 2 - 2sin^2 x`. Proved.

Answer 2

`LHS=cos4x+cos2x`

`=1-2sin^2\x+1-2sin^2x`

`=2-2sin^2\x-2sin^2x=RHS`

[ using formula `cos2theta=1-2sin^2theta]`