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Prove with Euler formula that $$ \cos(x-y) = \cos(x)\cos(y) - \sin(x)\sin(y). $$

I know how to find $\cos(x+y)$, but as for $\cos(x-y)$, I'm clueless.

Thanks.

Daniel W. Farlow
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Lou
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    Everywhere in your calculation for $\cos(x+y)$ replace $y$ by $-y$. – André Nicolas Jan 30 '15 at 02:42
  • see the answer to this question here http://math.stackexchange.com/questions/1108447/proving-sine-of-sum-identity-for-all-angles/1108520#1108520 – abel Jan 30 '15 at 02:55

2 Answers2

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$e^{i(x-y)}=e^{ix}e^{-iy}=(\cos x+i\sin x)(\cos y-i\sin y)=\cos x\cos y-i\cos x\sin y+i\sin x \cos y+\sin x\sin y$.

$e^{-i(x-y)}=e^{-ix}e^{iy}=(\cos x-i\sin x)(\cos y+i\sin y)=\cos x\cos y+i\cos x\sin y-i\sin x \cos y+\sin x\sin y$.

Then $\cos(x-y)=\dfrac{e^{i(x-y)}+e^{-i(x-y)}}{2}=\dfrac{2(\cos x\cos y + \sin x\sin y)}{2}=\cos x\cos y + \sin x\sin y$

jimbo
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\begin{align*} e^{ix} & = \cos x+ i \sin x\\ e^{i(x-y)} & = \frac{e^{ix}}{e^{iy}}\\ \frac{e^{ix}}{e^{iy}} & =\frac{\cos x + i \sin x}{\cos y+i\sin y}\\ \frac{\cos x + i \sin x}{\cos y+i\sin y} & =\frac{\cos x + i \sin x}{\cos y+i\sin y}\frac{\cos y - i \sin y}{\cos y -i\sin y}\\ & =\frac{\cos x \cos y+i\sin x\cos y-i\cos x\sin y+\sin x \sin y}{\cos^2 y+ \sin^2 y}\\ & ={\sin x\sin y + \cos x \cos y+i(\cos x \sin y -\sin x \cos y)} \end{align*}

Equating real and imaginary parts: $$\boxed{\cos [x-y]=\sin x\sin y + \cos x \cos y \text{ and } \sin[x-y]=\cos x \sin y -\sin x \cos y}$$

Teoc
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