If $u=(1+\cos t)(1+\cos 2t)-\sin t\sin 2t$ and $v=\sin t(1+\cos 2t)+\sin 2t(1+\cos t)$, then $u^2+v^2=4(1+\cos t)(1+\cos 2t)$ 
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.

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 
$$16\cos^2\frac{\theta}{2}\cos^2\theta$$
