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How would you prove $\sin^2x + \cos^2x = 1$ using Euler's formula?

$$e^{ix} = \cos(x) + i\sin(x)$$

This is what I have so far:

$$\sin(x) = \frac{1}{2i}(e^{ix}-e^{-ix})$$

$$\cos(x) = \frac{1}{2} (e^{ix}+e^{-ix})$$

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You're very close! Just square each of the two equations you have, and add them. Then watch magic happen. –  Arthur Feb 28 '13 at 15:36
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In general, $$\frac{1}{4}(a+b)^2 - \frac{1}{4}(a-b)^2 = ab$$ –  Thomas Andrews Feb 28 '13 at 15:38
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3 Answers 3

up vote 7 down vote accepted

Multiply $\mathrm e^{\mathrm ix}=\cos(x)+\mathrm i\sin(x)$ by the conjugate identity $\overline{\mathrm e^{\mathrm ix}}=\mathrm e^{-\mathrm ix}=\cos(x)-\mathrm i\sin(x)$.

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$e^{ix}=\cos x+i\sin x$

Putting $x=y$ and $x=-y$ respectively, $ e^{iy}=\cos y+i\sin y$ and $e^{-iy}=\cos(-y)+i\sin(-y)=\cos y-i\sin y$

So, $(\cos y+i\sin y)(\cos y-i\sin y)=e^{iy}e^{-iy}$

$\implies \cos^2y+\sin^2y=1$

or, $$(\cos y)^2+(\sin y)^2$$ $$=\left(\frac{e^{iy}+e^{iy}}{2}\right)^2+\left(\frac{e^{iy}-e^{-iy}}{2i}\right)^2$$ $$=\frac{(e^{iy}+e^{iy})^2-(e^{iy}-e^{iy})^2}4=\frac{4e^{iy}e^{-iy}}4=\frac44$$

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Why are you messing around with $y$? That seems an unnecessary complication to the answer. –  Thomas Andrews Feb 28 '13 at 15:39
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@ThomasAndrews, as I'm putting $x=y$ and $x=-y$. Putting $x=-x$ may lead to confusion –  lab bhattacharjee Feb 28 '13 at 15:41
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you just need to finish it.

$\sin^2(x) = \frac{1}{-4} ((e^{ix})^2 - 2 e^{ix}e^{-ix} + (e^{-ix})^2)$

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

$\sin^2(x) + \cos^2(x) = \frac{1}{4} 4 e^{ix}e^{-ix} = 1$

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