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Favorite Old Threads - Best Math Thread # 2

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Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Originally posted by tracker10 on 2/2/2011.

If $\sin(A)+\sin(B)=x$ and $\cos(A)+\cos(B)=y$, then prove that
$$\sin(A+B)=\frac{2xy}{x^{2}+y^{2}}.$$

My solution:

$$ x= \sin(A)+ \sin(B)=2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right),$$ and similarly

$$ y= \cos(A)+ \cos(B)=2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right).$$

Therefore, we have that

$$\frac{2xy}{x^{2}+y^{2}}=\frac{8 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A+B}{2}\right) \cos^{2}\left(\frac{A-B}{2}\right)}{4 \sin^{2}\left(\frac{A+B}{2}\right) \cos^{2}\left(\frac{A-B}{2}\right)+4 \cos^{2}\left(\frac{A+B}{2}\right) \cos^{2}\left(\frac{A-B}{2}\right)}$$
$$=\frac{8 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A+B}{2}\right) \cos^{2}\left(\frac{A-B}{2}\right)}{4 \cos^{2}\left(\frac{A-B}{2}\right)}$$
$$=2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A+B}{2}\right)$$
$$= \sin(A+B).$$

QED.