If
$$u = (1+\cos\theta)(1+\cos2\theta) – \sin\theta \sin 2\theta \qquad v = \sin\theta (1+\cos2\theta) + \sin2\theta(1+\cos\theta)$$
then show that
$$u^2 + v^2 = 4(1+\cos\theta)(1+\cos2\theta)$$
I have simplified the values of $u$ and $v$ and got:
$$u = 2\cos\theta (1+\cos2\theta) \qquad v=\sin2\theta(1+\cos\theta)$$
Then I tried to square both $u$ and $v$ individually before doing summation.
Still could not prove the statement.
Best Answer
Assigning $a$, $b$, $c$, $d$ as $$a := 1+\cos\theta \qquad b := 1+\cos2\theta \qquad c := \sin\theta \qquad d := \sin 2\theta$$
we can write $$u = ab - c d \qquad v = bc + ad$$ When squaring and adding, the $-2abcd$ will cancel with $2abcd$, leaving $$u^2 + v^2 = a^2 b^2 + c^2 d^2 + b^2 c^2 + a^2 d^2 = \left(a^2+c^2\right)\left(b^2+d^2\right)$$
Now, $$\begin{align} a^2 + c^2 &= \left(1 + 2 \cos\theta + \cos^2\theta\right) + \sin^2\theta = 2\left(1 + \cos\theta\right) \\[4pt] b^2 + d^2 &= \left(1 + 2 \cos 2\theta + \cos^22\theta\right) + \sin^22\theta = 2\left(1 + \cos 2\theta\right) \end{align}$$ and the result follows. $\square$
Note. The sum reduces even further to $$16\cos^2\frac{\theta}{2}\cos^2\theta$$