What is the quadratic trig identity for cosine when simplified?

[karush]

$$\cos\left({4x}\right)
=8\sin^4\left({x}\right)
-8\sin^2\left({x}\right)
+1$$

I thought this would break down
nice from the quadratic but it didn't.

 

[MarkFL]

First, apply a double-angle identity for cosine:

\(\displaystyle \cos(4x)=1-2\sin^2(2x)\)

Next, apply the double-angle identity for sine:

\(\displaystyle \cos(4x)=1-2\left(2\sin(x)\cos(x)\right)^2=1-8\sin^2(x)\cos^2(x)\)

Can you proceed? :D

 

[karush]

$$1-8\sin^4\left({x}\right)+8\sin^2\left({x}\right)$$
=1-8 (\sin^2\left({x}\right)(\sin^2\left({x}\right)-1))$$

 

[MarkFL]

$$1-8\sin^2\left({x}\right)
\cos^2\left({x}\right)
=1-8 (\sin^2\left({x}\right)(\sin^2\left({x}\right)-1))$$

Not quite:

\(\displaystyle \sin^2(x)+\cos^2(x)=1\implies\cos^2(x)=1-\sin^2(x)\)

 

[karush]

$$1-8\left[\sin^2\left({x}\right)\left(1-\sin^2\left({x}\right)\right)\right]$$
$$1-8\sin^2\left({x}\right)+8\sin^4\left({x}\right)$$