Given that sin2x = 2sinxcosx, deduce that cos2x = cos^2x - sin^2x

None of the other answerers appear to have actually answered the question posed.

My answer below repeatedly uses the "well-known" trig identity sin^2 x + cos^2 x = 1
Start with
sin 2x = 2sin x cos x
Square both sides
sin^2 (2x) = 4 (sin^2 x)(cos^2 x)
Now cos^2 (2x) = 1 - sin^2(2x)
= 1 - 4(sin^2 x)(cos^2 x)
= 1 - 4(sin^2 x)(1 - sin^2 x)
= 1 - 4(sin^2 x) + 4 sin^4 x
cos^2 (2x) = (1 -2 sin^2 x)^2
Square root both sides
cos (2x) = 1 - 2 sin^2 x
= sin^2 x + cos^2 x -2 sin^2 x
= cos^2 x - sin^2 x

cos ( x + x ) = cos x cos x - sin x sin x
cos 2x = cos ² x - sin ² x

cos (A+B) = cos A cos B - sin A sin B

let A = B = x

cos 2x
= cos x cos x - sin x sin x
= cos^2 x - sin ^2 x

why should i ?