Question

Prove that: cos $4x=1-8sin^{2} x cos^{2}x$

Answer 1

hello ☺

cos4x as cos2(2x)
=1-2sin2(2x)

 =1-2(2sinx . cosx)2                         
{ sin2x = 2sinx. cosx}

 =1-2(4sin2x.cos2x)

 =1-8sin2x.cos2x

LHS=RHS

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cos4x = cos2(2x)

2cos^2(2x) - 1

2(cos2x)^2 -1 

2[(2cosx -1 )]^2 -1

2[(4cos^2(x)+1-4cox)] -1

8cos^2(x) - 8cosx +1