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$\cos2a=\cos^2a-\sin^2a$

I know that euler's formula is $e^{ix} = \cos(x) + i\sin(x)$

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

where would I go from here?

Sebastiano
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Xiào
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    Welcome to Mathematics Stack Exchange. Here is a MathJax tutorial. Did you mean $(e^{i\color{red}a})^2=\cos^2(a)\color{red}-\sin^2(a)+2i\sin(a)\cos(a)$? – J. W. Tanner Sep 04 '23 at 20:29
  • Yes I did! Thank you, ill fix that – Xiào Sep 04 '23 at 20:33
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    From there, you could take the real part of both sides. Cf. this – J. W. Tanner Sep 04 '23 at 20:46
  • @J.W.Tanner In my opinion, the offered proofs of the angle sum identities have the defect of assuming that the angles $~\alpha, \beta, ~$ and $~(\alpha + \beta)~$ are all acute angles. Agreed that consideration of obtuse angles is ugly; however it is necessary and I have never seen any attempt to consider the obtuse angles. – user2661923 Sep 04 '23 at 21:56

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