This question is from S.L.LONEY- If $\tan(45°+\frac{y}{2})=\tan^3(45°+\frac{x}{2})$, prove that $\frac{\sin y}{\sin x}=\frac{3+\sin^2x}{1+3\sin^2x}$. I don't know what to do. I am getting nasty expressions but nothing in $\sin x$. Thanks.
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$$\cos2(45^\circ+y/2)=\dfrac{1-\tan^2(45^\circ+y/2)}{1+\tan^2(45^\circ+y/2)}$$
$$\iff-\sin y=\dfrac{1-\tan^6(45^\circ+x/2)}{1+\tan^6(45^\circ+x/2)}$$
Now $$\tan^6(45^\circ+x/2)=\left[\tan^2(45^\circ+x/2)\right]^3$$
and $$\tan^2(45^\circ+x/2)=\dfrac{1-\cos2(45^\circ+x/2)}{1+\cos2(45^\circ+x/2)}=\dfrac{1+\sin x}{1-\sin x}$$
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