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.
 A: 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 
$$16\cos^2\frac{\theta}{2}\cos^2\theta$$
A: Let $f(\theta)=1+\cos\theta+i\sin\theta$.
Then $|f(\theta)|^2=(1+\cos\theta)^2+\sin^2\theta=2(1+\cos\theta)$.
Note that $f(\theta)f(2\theta)=u+iv$.
So, $u^2+v^2=|f(\theta)f(2\theta)|^2=|f(\theta)|^2|f(2\theta)|^2=4(1+\cos\theta)(1+\cos2\theta)$.
A: $$u^2+v^2=(1+\cos\theta+\cos2\theta+\cos3\theta)^2+(\sin\theta+\sin2\theta+\sin3\theta)^2=$$
$$=4+2(\cos\theta+\cos2\theta+\cos3\theta)+2(\cos\theta\cos2\theta+\sin\theta\sin2\theta)+$$
$$+2(\cos\theta\cos3\theta+\sin\theta\sin3\theta)+2(\cos3\theta\cos2\theta+\sin3\theta\sin2\theta)=$$
$$=4+2(\cos\theta+\cos2\theta+\cos3\theta)+2(\cos\theta+\cos2\theta+\cos\theta)=$$
$$=4+6\cos\theta+4\cos2\theta+2\cos3\theta=4+4\cos\theta+4\cos2\theta+4\cos\theta\cos2\theta=$$
$$=4(1+\cos\theta)(1+\cos2\theta).$$
A: Simplifying $u$ we get
$$u=2\,\sin \left( x \right) \cos \left( x \right)  \left( 2\,\cos \left( 
x \right) +1 \right) 
$$ and $v$
$$v=2\,\sin \left( x \right) \cos \left( x \right)  \left( 2\,\cos \left( 
x \right) +1 \right) 
$$ so
$$u^2+v^2=8\, \left( \cos \left( x \right)  \right) ^{2} \left( 1+\cos \left( x
 \right)  \right) 
$$
and the right-hand side:$$8\, \left( \cos \left( x \right)  \right) ^{2} \left( 1+\cos \left( x
 \right)  \right) 
$$
which is clearly the same.
A: You simplified the values incorrectly. See below. 
Alternatively:
$$\begin{align}\color{blue}u&=(1+\cos\theta)(1+\cos2\theta) - \sin\theta \sin 2\theta=\\
&=1+\cos \theta +\cos 2\theta +\cos \theta \cos 2\theta-\sin\theta \sin2\theta=\\
&=1+\cos\theta+\cos2\theta+\cos3\theta=\\
&=1+\cos2\theta+2\cos\theta\cos2\theta= \ \ \ \ [\color{red}{\ne 2\cos\theta (1+\cos2\theta)}]\\
&=\color{blue}{1+\cos2\theta(1+2\cos\theta)};\\
\color{green}v&=\sin\theta (1+\cos2\theta) + \sin2\theta(1+\cos\theta)=\\
&=\sin\theta+\sin2\theta+\sin\theta\cos2\theta+\cos\theta\sin2\theta=\\
&=\sin\theta+\sin2\theta+\sin3\theta=\\
&=\sin2\theta+2\sin2\theta\cos\theta=\\
&=\color{green}{\sin2\theta(1+2\cos\theta)}; \ \ \ \ \ \ \ \ [\color{red}{\ne \sin2\theta(1+\cos\theta)}]\\
\color{blue}{u^2}+\color{green}{v^2}&=\color{blue}{1+2\cos2\theta(1+2\cos\theta)+\cos^22\theta(1+2\cos\theta)^2}+\color{green}{\sin^22\theta(1+2\cos\theta)^2}=\\
&=1+2\cos2\theta(1+2\cos\theta)+1\cdot(1+2\cos\theta)^2=\\
&=1+2\cos2\theta+4\cos\theta\cos2\theta+1+4\cos\theta+4\cos^2\theta=\\
&=2+2\cos2\theta+4\cos\theta\cos2\theta+4\cos\theta+2(\cos2\theta+1)=\\
&= 4(1+\cos\theta)(1+\cos2\theta).\end{align}$$
