<s>[INST] Prove that $\sin \theta=\frac{3 \sin \alpha+\sin^3 \alpha}{1+3\sin^2 \alpha}$ using given condition If $$\tan(\frac{\pi}{4}+\frac{\theta}{2})=\tan^3(\frac{\pi}{4}+\frac{\alpha}{2})$$, then prove that $$\sin \theta=\frac{3 \sin \alpha+\sin^3 \alpha}{1+3\sin^2 \alpha}$$
I tried using the fact that
$\frac{\cos A}{1-\sin A}=\tan(\frac{\pi}{4}+\frac{A}{2})$ but now not able to eliminate $\cos \alpha$ and $\cos \theta$. How should I proceed?
 [/INST] 
Notations and assumptions

$$
a=\tan{\theta\over2},\quad b=\tan{\alpha\over2},\quad p=\sin{\theta},\quad q=\sin\alpha
$$

Consequence of our notation

$$
{1+a\over1-a}=\left({1+b\over1-b}\right)^3\cdots\spadesuit
$$

An identity

$$
\sin z=2\sin{z\over2}\cos{z\over2}=2\tan{z\over2}\cos^2{z\over2}={2\tan{z\over2}\over\sec^2{z\over2}}={2\tan {z\over2}\over1+\tan^2{z\over2}}\cdots\heartsuit
$$

Using $\heartsuit$  and 

$$
{x\over y}={u\over v}\implies {y+x\over y-x}={v+u\over v-u}\cdots\diamondsuit
$$
we get to
$$
{1+\sin z\over1-\sin z}=\left({1+\tan{z\over2}\over1-\tan{z\over2}}\right)^2\cdots\clubsuit
$$

Using $\clubsuit$ we get

$$
{1+p\over1-p}=\left({1+a\over1-a}\right)^2,\quad
{1+q\over1-q}=\left({1+b\over1-b}\right)^2$$

Using our $\spadesuit$ we get

$${1+p\over1-p}=\left({1+q\over1-q}\right)^3
$$

Using $\diamondsuit$ we get the required identity

$$
-{1\over p}={(1-q)^3+(1+q)^3\over(1-q)^3-(1+q)^3}\implies p={(1+q)^3-(1-q)^3\over(1+q)^3+(1-q)^3}={3q+q^3\over1+3q^2}
$$
</s>