If `tan(pi/4+theta/2)=tan^3(pi/4+alpha/2)` then `sin(theta)=(sin(alpha)(3+sin^2(alpha)))/(1+3sin^2(alpha))`

Explanation

We know that,

`tan(pi/4+alpha)=(cosalpha+sinalpha)/(cosalpha-sinalpha) `

`∴tan^2(pi/4+alpha)=(1−sin2alpha)/(1+sin2alpha)`


Squaring the given relation and applying componendo and dividendo we get,

`=>(1+sintheta)/(1-sintheta)`

`=>(1+sinalpha)^3/(1-sinalpha)^3`

`=>(2sintheta)/2=(2(3sinalpha+sin^3alpha) )/(2(1+3sin^2alpha))`

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