Trigonometry Identity (Proof): $ \sin^4\theta +\cos^4 \theta =1-2\sin^2 \theta \cos^2 \theta $ 

Question: Prove that  $$ \sin^4\theta +\cos^4 \theta =1-2\sin^2 \theta \cos^2 \theta $$


What I have attempted (Usually I start of with the complex side)
So starting with the LHS
$$ \sin^4\theta +\cos^4 \theta =1-2\sin^2 \theta \cos^2 \theta $$ 
$$ (\sin^2\theta)^2 + \cos^4 \theta =1-2\sin^2 \theta \cos^2 \theta $$ 
$$ (1-\cos^2\theta)^2 + \cos^4 \theta =1-2\sin^2 \theta \cos^2 \theta $$ 
$$ (1-\cos^2\theta)(1-\cos^2\theta) + \cos^4 \theta =1-2\sin^2 \theta \cos^2 \theta $$ 
$$ 1 - 2\cos^2\theta + \cos^4\theta + \cos^4\theta =1-2\sin^2 \theta \cos^2 \theta $$
$$ 1 - 2\cos^2\theta + 2\cos^4\theta  =1-2\sin^2 \theta \cos^2 \theta $$
Now I am stuck... is my approach correct?

Alternatively, you may just square the identity

$$
\cos^2\theta +\sin^2 \theta=1
$$ 

giving
$$
\cos^4\theta+2\cos^2\theta\sin^2\theta +\sin^4 \theta=1.
$$
