How do I simplify these trig functions so that the right and left sides are the same?
you could note that both sides are $\displaystyle \cos 2x$, but that would be too easy.
instead, use the fact that $\displaystyle \sin^2 x + \cos^2 x = 1$. start with the left hand side of the equation and use the identity i typed abpve to replace either the sine or cosine to get the right hand side
no, it's not correct. you still did not do what i asked.
$\displaystyle \cos^2 x - \sin^2 x = 1 - 2\sin^2 x$
consider the LHS
$\displaystyle \cos^2 x - \sin^2 x = \left( 1 - \sin^2 x \right) - \sin^2 x$ ...........by the identity i showed you
$\displaystyle = 1 - 2 \sin^2 x$
$\displaystyle = RHS$
or, we could consider the RHS:
$\displaystyle 1 - 2 \sin^2 x = \sin^2 x + \cos^2 x - 2 \sin^2 x$
$\displaystyle = \cos^2 x = \sin^2 x$
$\displaystyle = LHS$
either way, we see that $\displaystyle \cos^2 x - \sin^2 x = 1 = 2 \sin^2 x$
but you could just as easily simplify both sides to $\displaystyle \cos 2x$