Prove the identity sin^2(x)-cos^2(x) = 2sin^2(x)-1

Since sin^2(x) + cos^2(x) = 1, we see that:
cos^2(x) = 1 - sin^2(x).

Substituting this in for cos^2(x) gives:
LHS = sin^2(x) - cos^2(x)
= sin^2(x) - [1 - sin^2(x)]
= sin^2(x) - 1 + sin^2(x), by distributing the negative
= 2sin^2(x) - 1
= RHS.

I hope this helps!

cos^2(x) + Sin^2 (x) =1 so cos^2(x)= 1-sin^2(X)
LHS of your question= sin^2(x)-(1-sin^2(x) )=sin^2(X)-1+sin^2(x)
=2sin^2(x)-1=same as RHS
LHS=lefthand side
RHS=righthand side

sin^2(x) + cos^2(x) = 1

Rearrange that to solve for cos^2(x). Plug it into the left hand side instead of cos^2(x).

cos^2x = 1 - sin^2x

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

:)>

sin^2(x)-cos^2(x) =
sin^2(x) - [1-sin^2(x)] = 2sin^2(x) - 1

umm what