Trigonometric equality proof $ \cos^2(\omega t) + \cos^2(\omega t + \delta) = \sin^2\delta + 2\cos(\omega t + \delta)\cos(\omega t)\cos(\delta)$ Looking to prove
$x = A\cos(\omega t)\\
y = A\cos(\omega t + \delta)
\\
\\$
YIELDS
$x^2-2xy\cos(\delta)+y^2=A^2\sin^2(\delta)$
Specifically we're trying to express the equations without any reference to $t$. If it helps, we can look in terms of ${x^2 + y^2 \over {A^2}}$
and just prove
$$ \cos^2(\omega t) + \cos^2(\omega t + \delta) = \sin^2\delta + 2\cos(\omega t + \delta)\cos(\omega t)\cos(\delta)$$
I know the angle sum identity $$\cos(\omega t + \delta) = \cos(\omega t) \cos(\delta) - \sin(\omega t)\sin(\delta)$$
$$ \cos^2(\omega t) + \cos^2(\omega t + \delta) \\ =\cos^2(\omega t) + [\cos(\omega t )\cos(\delta) - \sin(\omega t)\sin(\delta)]^2\\ 
=\cos^2(\omega t) + \cos^2(\omega t )\cos^2(\delta)+ \sin^2(\omega t) \sin^2(\delta)-2\sin(\omega t)\sin(\delta)\cos(\omega t )\cos(\delta) $$
But I can't get anywhere after that. I suspect that there is some other trig identity I either don't know or am overlooking somewhere.
This is from a physics textbook talking about two dimensional oscillation with the same $\omega$ but offset by a $\delta$ phase angle. It's not a problem i'm just trying to follow along the text and fill in the author's gaps.

By setting $\omega t = x, \omega t + \delta = y$, it is equivalent to showing:
$\cos^2 x + \cos^2 y = \sin^2(y-x) + 2\cos y\cos x \cos (y-x) $
$\cos^2 x + \cos^2 y  - \sin^2(y-x) \\= \cos^2 x + \cos^2 y - (\sin y\cos x - \cos y\sin x)^2 \\=  \cos^2 x + \cos^2 y - \sin^2 y\cos^2 x - \cos^2 y\sin^2 x + 2\sin y\cos x \cos y\sin x \\=  \cos^2 x (1- \sin^2 y) + \cos^2 y (1-\sin^2 x) + 2\sin y\cos x \cos y\sin x \\= \cos^2 x \cos^2 y + \cos^2 y \cos^2 x + 2\sin y\cos x \cos y\sin x \\=2\cos x\cos y(\cos x \cos y + \sin x\sin y) \\= 2\cos x\cos y\cos (y-x)$
from which the result immediately follows.
