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I need help with understanding how one can rewrite:

$\sin\alpha\cos\beta$

to be equal to: $\frac{1}{2}(\sin (\alpha + \beta) + \sin(\alpha-\beta))$ using Eulers formula.

I know that it probably is quite simple but I cannot get my head around this... Frustrating!

Eulers formulas:

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

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

Thank you kindly for your help!

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4 Answers 4

up vote 2 down vote accepted

$$\begin{align*} \sin\alpha\cos\beta&=\frac1{4i}(e^{i\alpha}-e^{-i\alpha})(e^{i\beta}+e^{-i\beta})\\ &=\frac1{4i}\Big(e^{i\alpha}e^{i\beta}+e^{i\alpha}e^{-i\beta}-e^{-i\alpha}e^{i\beta}-e^{-i\alpha}e^{-i\beta}\Big)\\ &=\frac1{4i}\Big(e^{i\alpha}e^{i\beta}-e^{-i\alpha}e^{-i\beta}+e^{i\alpha}e^{-i\beta}-e^{-i\alpha}e^{i\beta}\Big)\\ &=\frac1{4i}\Big(e^{i(\alpha+\beta)}-e^{-i(\alpha+\beta)}+e^{i(\alpha-\beta)}-e^{-i(\alpha-\beta)}\Big)\\ &=\frac12\Big(\sin(\alpha+\beta)+\sin(\alpha-\beta)\Big) \end{align*}$$

I actually got this by working from each end towards the middle, however.

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Thanks for the excellent answer! –  Lukas Arvidsson Feb 19 '13 at 15:32
    
@Lukas: You’re welcome! –  Brian M. Scott Feb 19 '13 at 15:35

Hint: You have the right formulas, so now try to FOIL their product and remember $$e^{i\alpha}\cdot e^{i\beta}=e^{i(\alpha+\beta)}\quad\text{and}\quad e^{i\alpha}\cdot e^{-i\beta}=e^{i(\alpha-\beta)}.$$

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An alternative proof is first showing the addition theorems: $$ \begin{align*}\cos(x+y) + i \sin(x+y)&= e^{i(x+y)}\\ &=e^{ix} \cdot e^{iy}\\ &= (\cos(x)+ i \sin(x)) \cdot (\cos(y)+ i \sin(y))\\ & = \cos(x)\cos(y)-\sin(x)\sin(y) + i( \cos(x)\sin(y)+\cos(y) \sin(x)) \end{align*}$$ Comparing real parts and imaginary parts give you both addition theorems. Now we take the right hand side $$\begin{align} \sin(x+y)+\sin(x-y)&=\cos(x)\sin(y)+\cos(y)\sin(x) + \cos(x)\sin(-y)+\cos(-y)\sin(x)\\ &= \cos(x)\sin(y)+\cos(y)\sin(x) - \cos(x)\sin(y) +\cos(y)\sin(x)\\&=2 \sin(x)\cos(y)\end{align}$$

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$(\sin\alpha)(\cos\beta) = \frac{1}{4i}(e^{i\alpha}-e^{-i\alpha})(e^{i\beta}+e^{-i\beta})$

by expanding the right hand side we get

$\frac{1}{4i}(e^{i\alpha+i\beta}-e^{-i\alpha-i\beta}+e^{i\alpha-i\beta}-e^{-i\alpha+i\beta}) = \frac{1}{4i}\big((e^{i(\alpha+\beta)}-e^{-i(\alpha+\beta)})+(e^{i(\alpha-\beta)} - e^{-i(\alpha-\beta)})\big)$

I think I'll stop here. =)

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