A great problem in Trigonometry

[DrunkenOldFool]

If $\sin x +\cos y=a$ and $\cos x+\sin y =b $, then what is $\tan\dfrac{x-y}{2}$ in terms of $a$ and $b$?

 

[sbhatnagar]

If $\sin x +\cos y=a$ and $\cos x+\sin y =b $, then what is $\tan\dfrac{x-y}{2}$ in terms of $a$ and $b$?

Hello DrunkenOldFool! We have the following equations

\[ \sin x +\cos y=a \\ \cos x+\sin y =b\]

Adding these two equations, we will get

\[ (\sin x +\sin y)+(\cos x + \cos y)=a+b\]

Note that $\sin x +\sin y = 2\sin \dfrac{x+y}{2} \cos \dfrac{x-y}{2}$ and $\cos x +\cos y = 2\cos \dfrac{x+y}{2} \cos \dfrac{x-y}{2}$.

\[ \begin{align*} \implies \ 2\sin \dfrac{x+y}{2} \cos \dfrac{x-y}{2}+2\cos \dfrac{x+y}{2} \cos \dfrac{x-y}{2}&=a+b \\ \implies 2\sin \dfrac{x+y}{2}+2\cos \dfrac{x+y}{2} &= \frac{a+b}{\cos \dfrac{x-y}{2}} \end{align*}\]

Multiply both sides by $\sin\dfrac{x-y}{2}$.

\[\begin{align*}2\sin\dfrac{x-y}{2}\sin \dfrac{x+y}{2}+2\sin\dfrac{x-y}{2}\cos \dfrac{x+y}{2} &= (a+b)\tan \dfrac{x-y}{2}\end{align*}\]

Apply trigonometric product to sum identities on the left hand side of the equation:

\[\begin{align*}\cos y- \cos x+\sin x -\sin y &= (a+b)\tan \dfrac{x-y}{2} \\ \underbrace{(\sin x +\cos y)}_{a}-\underbrace{(\sin y +\cos x)}_{b} &= (a+b)\tan \dfrac{x-y}{2} \\ \tan \dfrac{x-y}{2} &= \frac{a-b}{a+b}\end{align*}\]