How to Simplify This Trigonometric Equation Using Substitutions?

[Fred1230]

Returning if I have to show the effort, I came to this:
\frac{\sin4\alpha}{1+\cos4\alpha}\cdot\frac{\cos2\alpha}{1+\cos2\alpha}\cdot\frac{\cos\alpha}{1+\cos\alpha}=\tan\frac{\alpha}{2}.
=
\frac{\sin4\alpha}{\sin^2\alpha+cos^2\alpha+\cos4\alpha}\cdot\frac{(\sin^2\alpha+cos^2\alpha)-2sin^2\alpha}{\sin^2\alpha+cos^2\alpha+\cos2\alpha}\cdot\frac{\cos\alpha}{\sin^2\alpha+cos^2\alpha+\cos\alpha}=\frac{\sin\alpha^2}{\cos2\alpha}.
I don't know how to use substitutions

 

[Fred1230]

s=\sin\alpha and c=\cos\alpha

 

[BvU]

Returning if I have to show the effort

Returning from where ?
What is the complete problem statement ?

!

https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

$\$

 

[neilparker62]

Returning if I have to show the effort, I came to this:
\frac{\sin4\alpha}{1+\cos4\alpha}\cdot\frac{\cos2\alpha}{1+\cos2\alpha}\cdot\frac{\cos\alpha}{1+\cos\alpha}=\tan\frac{\alpha}{2}.
=
\frac{\sin4\alpha}{\sin^2\alpha+cos^2\alpha+\cos4\alpha}\cdot\frac{(\sin^2\alpha+cos^2\alpha)-2sin^2\alpha}{\sin^2\alpha+cos^2\alpha+\cos2\alpha}\cdot\frac{\cos\alpha}{\sin^2\alpha+cos^2\alpha+\cos\alpha}=\frac{\sin\alpha^2}{\cos2\alpha}.
I don't know how to use substitutions

Substitute $\alpha=60^{\circ}$ in your expression and check if you come out with $\tan30^{\circ}$. If not it's back to the drawing board!

 

[haruspex]

Returning if I have to show the effort, I came to this:
\frac{\sin4\alpha}{1+\cos4\alpha}\cdot\frac{\cos2\alpha}{1+\cos2\alpha}\cdot\frac{\cos\alpha}{1+\cos\alpha}=\tan\frac{\alpha}{2}.
=
\frac{\sin4\alpha}{\sin^2\alpha+cos^2\alpha+\cos4\alpha}\cdot\frac{(\sin^2\alpha+cos^2\alpha)-2sin^2\alpha}{\sin^2\alpha+cos^2\alpha+\cos2\alpha}\cdot\frac{\cos\alpha}{\sin^2\alpha+cos^2\alpha+\cos\alpha}=\frac{\sin\alpha^2}{\cos2\alpha}.
I don't know how to use substitutions

Do you know a formula for $\cos(2\alpha)$ in terms of $\cos(\alpha)$?