Question #02e7c

2 Answers
P dilip_k
Nov 11, 2016

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

Nghi N
Nov 28, 2016

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.

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