# TrigonometryA great problem in Trigonometry

#### DrunkenOldFool

##### New member
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

##### Active member
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*}

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