Question

Question #02e7c

Answer 1

Given

`cos4x = 1 - bsin^2x cos^2x`

`=>1-2sin^2 2x = 1 - b/4(2sinx cosx)2`

`=>2sin^2 2x = b/4sin^2 2x`

`=>b/4=2`

`=>b=8`

Answer 2

b = 8

Explanation:

`cos 4x =1 - bsin^2 x.cos^2 x` (1)
Use 2 trig identities:
`cos 2a = 1 - 2sin^2 a`, and
sin 2a = 2sin a.cos a

Replace a by x, we get:
`cos 4x = 1 - 2sin^2 (2x)`
Since sin 2x = (2sin x.cos x), then:
`2sin^2 (2x) = 2(2sinx.cos x)^2`
There for:
`cos 4x = 1 - 2(2sin x.cos x)^2 = 1 - 8sin^2 x.cos^2 x`. (2)
Compare (1) and (2), we get b = 8.