I came across this problem on varsity tutors. A part of the answer walk through states that $\sin^2(x)$ can equal $\cos^2(x) - 1$. This is stated more than once on the site. I do not see how this is possible.
I can see that \begin{align} \sin^2(x) &= 1 - \cos^2(x) \\ \cos^2(x) &= 1 - \sin^2(x) \end{align}
if $\cos^2(x)-1$ were to be the equation. then it would need to equal $-\sin^2(x)$.
Am I missing something?
You want to solve $\sin^2 x = \cos^2 x - 1$. Substitution via the known identity $\sin^2 x + \cos^2 x = 1$ yields $$\sin^2 x = (1-\sin^2 x) - 1 = -\sin^2 x,$$ which immediately implies that $\sin^2 x=0$, hence $\sin x = 0$. So the solutions are $x=n \pi$ for integer $n$.
$0\le\sin^2x=\cos^2x-1\le0$, therefore $\sin x=0$ and $x=k\pi$.
If $\sin^2(x)=\cos^2(x)-1$ then $$2\sin^2(x)=\sin^2(x)+\sin^2(x)=\sin^2(x)+(\cos^2(x)-1)=1-1=0,$$ and so $\sin^2(x)=0$. This implies that $x=k\pi$ for some integer $k$.
$\sin^2(x) = \cos^2(x)-1 \iff 1 = \cos^2(x)-\sin^2(x) = \cos(2x) \iff 2x = 2\pi n \iff x = n\pi$ for some $n \in \mathbb{Z}$.
When I was looking at your statement, I first looked at it as an identity, and then as an equation.
As an identity, no. It's well known that $\sin^2 x + \cos^2 x = 1$ and you're correct that $\cos^2 x - 1$ would yield $-\sin^2 x$, so that identity would be incorrect.
As an equation, yes. With manipulation of the identity $\sin^2 x + \cos^2 x = 1$, the equation yields $\sin x = 0$, to which $x = n \pi$.
