$ 1 - \cos 2 \Theta$ can be rewritten as $1 - \left( 1 - 2 \sin^2 \Theta\right)$ - I don't understand why though

Question

Going through a video I saw this and wasn't sure how to sort it -

given the following :

$$ r = 4 \left( 1 - \cos 2 \Theta \right) $$

the part in parenthesis $ 1 - \cos 2 \Theta$ can be rewritten as $1 - \left( 1 - 2 \sin^2 \Theta\right)$.

I'm not sure where this comes from though? I have a sheet of identities, but this isn't on there. If anyone could point me in the right direction as to why this can be rewritten as such that'd be ace.

Looking through a sheet such as this one I can't really intuit them, I just use them when needed (I couldn't write a proof).

Cheers.

Answer 1

There are three forms of the $\cos{2x}$ "double-angle" formula, which you can shift between by using $\cos^2{x}+\sin^2{x}=1$: $$ \cos{2x} = \cos^2{x}-\sin^2{x} = \cos^2{x}-(1-\cos^2{x}) = 2\cos^2{x}-1 \\ = (1-\sin^2{x})-\sin^2{x}, $$ so $$ \begin{align} \cos{2x} &= \cos^2{x}-\sin^2{x} \\ &= 2\cos^2{x}-1 \\ &= 1-2\sin^2{x}. \end{align} $$

Answer 2

Use this identity $\sin^2(x)=\frac{1}{2}(1-\cos(2x))$:

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$$ 1 - \cos(2x)=1 - \left( 1 - 2 \sin^2(x)\right) \Longleftrightarrow$$ $$ 1 - \cos(2x)=1 - \left( 1 - 2 \left(\frac{1}{2}(1-\cos(2x))\right)\right) \Longleftrightarrow$$ $$ 1 - \cos(2x)=1 - \left( 1 - \left(1-\cos(2x)\right)\right) \Longleftrightarrow$$ $$ 1 - \cos(2x)=1 - \left( 1 -1+\cos(2x)\right) \Longleftrightarrow$$ $$ 1 - \cos(2x)=1 - \left( 0+\cos(2x)\right) \Longleftrightarrow$$ $$ 1 - \cos(2x)=1 - \cos(2x) $$