Prove that $\frac {\sin X+\sin Y-\sin(X+Y)}{\sin X+\sin Y+\sin(X+Y)}=\tan\frac{X}{2}\tan\frac{Y}{2}$

Question

Prove that $\frac {\sin X+\sin Y-\sin(X+Y)}{\sin X+\sin Y+\sin(X+Y)}=\tan\frac{X}{2}\tan\frac{Y}{2}$

I tried expanding $\frac {\sin X+\sin Y-\sin(X+Y)}{\sin X+\sin Y+\sin(X+Y)}$ and expanding $\tan\frac{X}{2}\tan\frac{Y}{2}$ and then cross multiplying , but I can't seem to find the answer.

Answer 1

Write $t=\tan(x/2)$ and $u=\tan(y/2)$. Then $$\sin x=\frac{2t}{1+t^2},$$ $$\sin y=\frac{2u}{1+u^2}$$ and $$\sin(x+y)=\sin x\cos y+\cos x\sin y=\frac{2t(1-u^2)+2u(1-t^2)}{(1+t^2)(1+u^2)}.$$ Therefore $$\sin x+\sin y\pm\sin(x+y) =\frac{2t(1+u^2)+2u(1+t^2)\pm\left(2t(1-u^2)+2u(1-t^2)\right)}{(1+t^2)(1+u^2)}.$$ Now expand out both possibilities.

Answer 2

Start with the LHS, use sum-to-product on sin(X)+sin(Y), and the double angle formula on sin(X+Y), the expression is equal to $$ \frac{\sin(\frac{X+Y}{2})\cos(\frac{X-Y}{2})-\sin(\frac{X+Y}{2})\cos(\frac{X+Y}{2})}{\sin(\frac{X+Y}{2})\cos(\frac{X-Y}{2})+\sin(\frac{X+Y}{2})\cos(\frac{X+Y}{2})} $$ cancelling sin((X+Y)/2) and use sum-to-product again, $$ = \frac{\sin\frac{X}{2}\sin\frac{Y}{2}}{\cos\frac{X}{2}\cos\frac{Y}{2}} $$ which is the RHS.

Answer 3

\begin{align*}\frac{\sin x+\sin y-\sin(x+y)}{\sin x+\sin y+\sin (x+y)}&=\dfrac{2\sin\dfrac{x+y}{2}\cos\dfrac{x-y}{2}-2\sin\dfrac{x+y}{2}\cos \dfrac{x+y}{2}}{2\sin\dfrac{x+y}{2}\cos\dfrac{x-y}{2}+2\sin\dfrac{x+y}{2}\cos \dfrac{x+y}{2}}\\&=\dfrac{\cos\dfrac{x-y}{2}-\cos \dfrac{x+y}{2}}{\cos\dfrac{x-y}{2}+\cos \dfrac{x+y}{2}}\\&=\dfrac{-2\sin\dfrac{x}{2}\sin\dfrac{-y}{2}}{2\cos\dfrac{x}{2}\cos \dfrac{-y}{2}}\\&=\tan\dfrac{x}{2}\tan\dfrac{y}{2}.\end{align*}

Answer 4

Denote: $x=a+b, y=a-b$, then: $$\frac {\sin (a+b)+\sin (a-b)-\sin(2a)}{\sin (a+b)+\sin (a-b)+\sin(2a)}=\\ \frac{\sin a\cos b+\require{cancel}\cancel{\cos a\sin b}+\sin a\cos b-\require{cancel}\cancel{\cos a\sin b}-\sin (2a)}{\sin a\cos b+\require{cancel}\cancel{\cos a\sin b}+\sin a\cos b-\require{cancel}\cancel{\cos a\sin b}+\sin (2a)}=\\ \frac{2\sin a\cos b-2\sin a\cos a}{2\sin a\cos b+2\sin a\cos a}=\\ \frac{\cos b-\cos a}{\cos b+\cos a}=\\ \frac{-\sin \frac{b-a}{2}\sin \frac{b+a}{2}}{\cos \frac{b-a}{2}\cos \frac{b+a}{2}}=\\ \frac{-\sin (-\frac y2)\sin (\frac x2)}{\cos (-\frac y2)\cos (\frac x2)}=\\ \tan\frac{X}{2}\tan\frac{Y}{2}.$$