Euler's Formula

**Gitano** · April 21st, 2009, 08:15 PM

Hello, I have a test coming up this Friday and I am completely stuck on this test review question.

The question asks to derive the indentities using the euler's formula.

sin2X = 2sinXcosX

cos2X = cos^2X-sin^2X

note: The 'X' variable stands for theta

Any help or feedback would be greatly appreciated, thanks.

**Chris L T521** · April 21st, 2009, 08:26 PM

Originally Posted by Gitano

Hello, I have a test coming up this Friday and I am completely stuck on this test review question.

The question asks to derive the indentities using the euler's formula.

sin2X = 2sinXcosX

cos2X = cos^2X-sin^2X

note: The 'X' variable stands for theta

Any help or feedback would be greatly appreciated, thanks.

Recall that Euler's formula is $e^{i\theta}=\cos\theta+i\sin\theta$ . From this, we see that $e^{i\theta}-e^{-i\theta}=2i\sin\theta\implies\sin\theta=\frac{e^{i \theta}-e^{-i\theta}}{2i}$ and $e^{i\theta}+e^{-i\theta}=2\cos\theta\implies \cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}$

Therefore,

$\begin{aligned}\sin\!\left(2\theta\right)&=\frac{e ^{i(2\theta)}-e^{-i(2\theta)}}{2i}\\&=\frac{\left(e^{i\theta}+e^{-i\theta}\right)\left(e^{i\theta}-e^{-i\theta}\right)}{2i}\\&=2\left(\frac{e^{i\theta}+e ^{-i\theta}}{2}\right)\left(\frac{e^{i\theta}-e^{-i\theta}}{2i}\right)\\&=2\cos\theta\sin\theta\\&=2 \sin\theta\cos\theta\end{aligned}$

Can you try to do something similar with $\cos\!\left(2\theta\right)$ ??

**TheEmptySet** · April 21st, 2009, 08:28 PM

Originally Posted by Gitano

Hello, I have a test coming up this Friday and I am completely stuck on this test review question.

The question asks to derive the indentities using the euler's formula.

sin2X = 2sinXcosX

cos2X = cos^2X-sin^2X

note: The 'X' variable stands for theta

Any help or feedback would be greatly appreciated, thanks.

Eulers identity is

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

$(e^{i\theta})^2=e^{i2\theta}=\cos(2\theta) + i\sin(2\theta)$

but it also equals

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

Now setting the real and immaginary parts equal we get

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

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

**Gitano** · April 21st, 2009, 08:59 PM

You guys are awesome, thank you.

Although I still find it a bit confusing since euler's formula is a bit abstract, by looking at these solutions hopefully I will understand it by Friday, appreciate it.

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