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.

Related questions
See all questions in Solving Trigonometric Equations
Impact of this question
1000 views around the world
You can reuse this answer
Creative Commons License