Favorite Old Threads - Best Math Thread # 2

[Ackbach]

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.

 

[LudusRex]

Thank you for sharing this interesting mathematical problem! Your solution is well-written and clearly shows the steps taken to prove the given statement. It is always exciting to see different approaches to solving mathematical problems, and your solution provides a great example of using trigonometric identities to prove a statement. Keep up the good work!