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I have to use Euler's Formula to prove that:

$$\cos^2(\theta) = \frac{\cos(2\theta)+1}{2}.$$

I have managed to prove this using trigonometric identities but I'm not sure how to use Euler's Formula or how it links into the question.

My method so far has been:

$$\frac{(\cos(2\theta)+1)}{2} = \frac{(\cos^2(\theta) - \sin^2(\theta)+1)}{2}$$

since

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

So $$\frac{(\cos(2\theta)+1)}{2} =\frac{2\cos^2(\theta)}{2} =\cos^2(\theta).$$

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Eulers identity $e^{i\theta} = \cos \theta + i\sin\theta$

$e^{i\theta} + e^{-i\theta} = 2\cos \theta\\ \frac 14 (e^{i\theta} + e^{-i\theta})^2 = \cos^2 \theta\\ \frac 14 (e^{2i\theta} + e^{-2i\theta} + 2) = \cos^2 \theta\\ \frac 14 (2\cos 2\theta + 2) = \cos^2 \theta\\ \frac 12 (\cos 2\theta + 1) = \cos^2 \theta$

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By Euler's formula $2\cos x=e^{ix}+e^{-ix}$

$$2\cos2\theta=(e^{i\theta}+e^{-i\theta})^2-2e^{i\theta}\cdot e^{-i\theta} =(2\cos\theta)^2-2$$

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Euler's formula can be used to prove the addition formula for both sines and cosines as well as the double angle formula (for the addition formula, consider $\mathrm{e^{ix}}$. The same method as the double angle formula works).

From Euler's formula, $\mathrm{e^{2ix}=cos(2x)+isin(2x)=(e^{ix})(e^{ix})=(cos(x)+isin(x))(cos(x)+isin(x))}$

Expand the brackets and compare the real and imaginary parts for the double angle formulas.

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$$\cos x = \frac{e^{ix} +e^{-ix}}{2}$$ Assume $e^{ix}=u$ $$\left(\cos x\right)^2 = \left(\frac{u +\frac{1}{u}}{2}\right)^2$$ $$\cos^2 x=\frac{u^2 +\frac{1}{u^2}}{4} +\frac{2}{4}$$ $$\cos^2 x = \frac{e^{2ix} +e^{-2ix}}{4 }+ \frac{1}{2}$$

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Start with Euler's formula:

${e}^{i\theta}=\cos{\theta}+i\sin{\theta}$

Squaring both sides gives:

${({e}^{i\theta})}^{2}={(\cos{\theta}+i\sin{\theta})}^{2}$

Which, using the laws of exponents and the expansion of brackets, becomes:

${e}^{2i\theta}=\cos^{2}{\theta}+2i\sin{\theta}\cos{\theta}+{i}^{2}\sin^{2}{\theta}$

The left can be written with the exponent as a multiple of $i$ and the right can be simplified because $i^{2}=-1$:

${e}^{i(2\theta)}=\cos^{2}{\theta}-\sin^{2}{\theta}+2i\sin{\theta}\cos{\theta}$

However as $e^{i(2\theta)}$ is in the form $e^{ix}$ it can be expressed through Euler's formula:

$\cos{2\theta}+i\sin{2\theta}=\cos^{2}{\theta}-\sin^{2}{\theta}+2i\sin{\theta}\cos{\theta}$

The real parts can be equated:

$\Re{(\cos{2\theta}+i\sin{2\theta})}=\Re{(\cos^{2}{\theta}-\sin^{2}{\theta}+2i\sin{\theta}\cos{\theta})}$

Which leaves:

$\cos{2\theta}=\cos^{2}{\theta}-\sin^{2}{\theta}$

As $\sin^2{\theta}=1-\cos^2{\theta}$:

$\cos{2\theta}=\cos^{2}{\theta}-(1-\cos^2{\theta})$

This bracket can be multiplied out to give:

$\cos{2\theta}=2\cos^{2}{\theta}-1$

Which can be rearranged to:

$\cos^{2}{\theta}=\frac{\cos{2\theta}+1}{2}$

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