Question

cos 2theta = 1 - 2sin^2theta

Answer 1

Hi ,

Here I am using A instead of theta.

LHS = cos2A

= cos( A + A )

= cosAcosA - SinA SinA

= cos² A - sin² A

= ( 1 - sin² A ) - sin² A

= 1 - sin² A - sin² A

= 1 - 2sin²A

= RHS

I hope this helps you.

:)

Answer 2

(I am taking value of theta to be x for convenience)

Consider splitting up cos2 theta into cos(x+x) form.

Then apply formula for cos(A+B)
which is CosA.CosB - SinA.SinB

Cos(x+x) = cos(x)*cos(x) - sin(x)*sin(x)

Cos 2x = cos^2 (x) - sin^2 (x)

= 1- sin^2 (x) - sin^2 (x),

since sin^2x + cos^2 x = 1


= 1 - 2*sin^2 (x)

Hope this helps you