Simplifying a trigonometric expression

[Saitama]

Homework Statement

If $\displaystyle \tan\left(\frac{\pi}{4}+\frac{y}{2}\right)=\tan^3\left( \frac{\pi}{4}+\frac {x}{2} \right)$, prove that $$\frac{\sin y}{\sin x}=\frac{3+\sin^2 x}{1+3\sin^2x}$$

Homework Equations

The Attempt at a Solution

$$\tan\left(\frac{\pi}{4}+\frac{y}{2}\right)=\frac{1+\tan (y/2)}{1-\tan(y/2)}=\frac{\cos (y/2)+\sin (y/2)}{\cos (y/2)-\sin(y/2)}=\frac{1+\sin y}{\cos y}$$
Similarly,$$\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)=\frac{1+\sin x}{\cos x}$$
Plugging them,
$$\frac{1+\sin y}{\cos y}=\left(\frac{1+\sin x}{\cos x}\right)^3$$
Stuck here. Need a few hints to proceed further.

Any help is appreciated. Thanks!

 

[tiny-tim]

Hi Pranav-Arora!

$$\frac{\cos (y/2)+\sin (y/2)}{\cos (y/2)-\sin(y/2)}=\frac{1+\sin y}{\cos y}$$

That's not the only way of simplifying that, is it?

(and remember, you want all sins and no coss)

 

[Saitama]

Hi Pranav-Arora!


That's not the only way of simplifying that, is it?

(and remember, you want all sins and no coss)

$$\tan\left(\frac{\pi}{4}+\frac{y}{2}\right)=\frac{\sin(\pi/4+y/2)}{\sin(\pi/4-y/2)}$$
Would that help?

 

[tiny-tim]

no, i mean find another simplification for $$\frac{\cos (y/2)+\sin (y/2)}{\cos (y/2)-\sin(y/2)}$$

 

[Saitama]

no, i mean find another simplification for $$\frac{\cos (y/2)+\sin (y/2)}{\cos (y/2)-\sin(y/2)}$$

I think I still cannot follow your hint. Would the following work?
$$\frac{\cos (y/2)+\sin (y/2)}{\cos (y/2)-\sin(y/2)}=\sqrt{\frac{1+\sin y}{1-\sin y}}$$

 

[tiny-tim]

yes!

keep going!​

 

[Saitama]

yes!

keep going!​

Thanks a lot tiny-tim! I have proved it now.