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.
 A: 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.
A: Alternative approach, similar to Deepak's answer.
$\underline{\text{intermediate results}}$
Lemma 1: $~~\cos(a+b) \cos(b-a) = \cos^2(a) - \sin^2(b).$ 
Proof: 
$(\cos a \cos b - \sin a \sin b) \times (\cos a \cos b + \sin a \sin b)$ 
$= (\cos^2 a ~\cos^2 b) - (\sin^2 a ~\sin^2 b)$ 
$= [(\cos^2 a)(1 - \sin^2 b)] - [(1 - \cos^2 a)(\sin^2 b)]$ 
$= \cos^2 a - (\cos^2 a ~\sin^2 b) - \sin^2 b + (\cos^2 a ~\sin^2 b).$
Lemma 2: $~~\cos(a + b) + \cos(b - a) = 2(\cos a)(\cos b).$ 
This is routinely shown via the angle sum formulas.

Let $a = \omega t, ~b = \delta.$ 
Then the problem is to prove that 
$$\cos^2 a + \cos^2 (a + b) ~=~ \sin^2 b + 2\cos(a+b)\cos(a)\cos(b). \tag1 $$
Using Lemma 2, the RHS of (1) above may be re-expressed as 

$= \sin^2 b + 
~~\left\{2 \cos(a + b) \left(\dfrac{1}{2}\right) [\cos(a + b) + \cos (b - a)]\right\}$  
$= \sin^2 b + \cos^2 (a + b) + \cos(a + b)\cos (b - a).$
Using Lemma 1, the RHS may therefore be re-expressed as: 
$\sin^2 b + \cos^2 (a + b) + \cos^2 a - \sin^2 b.$
Therefore, the RHS of (1) equals the LHS of (1).
