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The problem is to prove that $\cos^4(\theta)-\sin^4(\theta)=cos(2\theta)$.

So, here are my steps so far:

$(\cos^2\theta)^2-(\sin^2\theta)^2=\cos(2\theta)$

$(\cos^2\theta+\sin^2\theta)(\cos^2\theta-\sin^2\theta)=\cos(2\theta)$

$(\cos^2\theta+\sin^2\theta)(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)=\cos(2\theta)$

I don't know where to go from here. Please help! I feel like I'm missing something super obvious but I don't know what...

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Well, $\cos^2(\theta)+\sin^2(\theta)=1$, and $\cos(2\theta)=\cos^2(\theta)-\sin^2(\theta)$ is a standard Double-Angle identity. Can you continue from there?

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  • $\begingroup$ Oh ya! I can't believe I forgot that....I got the answer. $\endgroup$
    – user680251
    Jun 9 '19 at 23:35
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There are a couple of nice ways we can do it. Firstly, note that: $$\cos^4(\theta)-\sin^4(\theta)=\bigl(\cos^2\theta-\sin^2\theta\bigr)\bigl(\cos^2\theta+\sin^2\theta\bigr)$$ and since: $$\cos^2\theta+\sin^2\theta=1\tag{1}$$ we can say that: $$\cos^4\theta-\sin^4\theta=\cos^2\theta-\sin^2\theta=\cos2\theta$$ which is the double angle formula for cosine. This can be proved geometrically a number of ways but also as shown below


We can also take advantage of De Moivre's theorem, which states that: $$\left(\cos x+i\sin x\right)^n=\cos(nx)+i\sin(nx)\tag{2}$$ Using the fact that: $$e^{ix}=\cos(x)+i\sin(x)\tag{3}$$ We can do this by saying: $$\cos(2x)+i\sin(2x)=\left(\cos x+i\sin x\right)^2$$ $$\cos(2x)+i\sin(2x)=\cos^2x+2i\cos x\sin x+i^2\sin^2x$$ $$\cos(2x)+i\sin(2x)=\cos^2x-\sin^2x+2i\cos x\sin x$$ Now if we separate the real and imaginary parts we are left with: $$\cos(2x)=\cos^2x-\sin^2x,\,\sin(2x)=2\sin(x)\cos(x)\tag{4}$$

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We have $$\cos^4(\theta)-\sin^4(\theta) = (\cos^2\theta+\sin^2\theta)(\cos^2\theta-\sin^2\theta) $$ where $(\cos^2\theta+\sin^2\theta) = 1$ and $(\cos^2\theta-\sin^2\theta) = 2\cos^2(\theta)-1 = \cos(2\theta)$

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Hint:$$\cos(t) = \frac{e^{it} + e^{-it}}{2}$$ and $$\sin(t) = \frac{e^{it} - e^{-it}}{2i}$$

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  • $\begingroup$ I just remembered the Double-Angle Identity. $\endgroup$
    – user680251
    Jun 10 '19 at 3:28

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