prove that 1-2sin^2 (π/4-x)=sin2x with explanations and rules please :) thank you in advance
cos 2x = 1 -2sin ^2x
1-2sin^2 (π/4-x)
= cos2(π/4-x)
= cos(π/2 - 2x)
= cos[-(2x - π/2)] => cosine is even function
= cos(2x - π/2) => co-function
= sin2x
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cos 2x = 1 -2sin ^2x
1-2sin^2 (π/4-x)
= cos2(π/4-x)
= cos(π/2 - 2x)
= cos[-(2x - π/2)] => cosine is even function
= cos(2x - π/2) => co-function
= sin2x