sin2x + cos4x=??
is there any one who can give the detailed solution of it....?????
Answers
Answered by
1
its sin2..or sin sq.??
vjrishabh01:
its sin not sin square
im 100% sure
Answered by
1
sin2x + cos4x = 2sin x.cos x + cos2(2x)
= 2sin x.cos x + 2(cos2x)^2 - 1
= 2sin x.cos x + 2{cos^x - sin^x}^2 - 1
= 2sin x.cos x + 2(cos^2x + sin^2x - 2sin x.cos x) - 1
= 2sin x.cos x + 2( 1 -2sin x. cos x) - 1
= 2sin x.cos x + 2 - 4sin x.cos x - 1
= 1 - 2sin x.cos x
= 1 - sin2x...........●
= 2sin x.cos x + 2(cos2x)^2 - 1
= 2sin x.cos x + 2{cos^x - sin^x}^2 - 1
= 2sin x.cos x + 2(cos^2x + sin^2x - 2sin x.cos x) - 1
= 2sin x.cos x + 2( 1 -2sin x. cos x) - 1
= 2sin x.cos x + 2 - 4sin x.cos x - 1
= 1 - 2sin x.cos x
= 1 - sin2x...........●
that means multiplication
Hello!! I can solve your question, if you ask it again.
Okk I am solving here.
If we take RHS = 2cos(π/4 + x).cos(π/4 - 3x)
= cos(π/4 + x + π/4 - 3x) + cos( π/4 + x - π/4 + 3x)...........................because {2cosx.cosy = cos(x+y) + cos(x-y)
= cos(π/2 - 2x) + cos(4x)
= sin2x + cos4x = L.H.S.......hence proved
Hope it was helpful. Please mark as brainliest.
ya buddy great used ur mind
Thankyou bro ! ! :)
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