How to Simplify This Trigonometric Equation Using Substitutions?

[Fred1230]

Returning if I have to show the effort, I came to this:
[tex]\frac{\sin4\alpha}{1+\cos4\alpha}\cdot\frac{\cos2\alpha}{1+\cos2\alpha}\cdot\frac{\cos\alpha}{1+\cos\alpha}=\tan\frac{\alpha}{2}.[/tex]
=
[tex]\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}.[/tex]
I don't know how to use substitutions

 

[Fred1230]

[tex]s=\sin\alpha[/tex] and [tex]c=\cos\alpha[/tex]

 

[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:
[tex]\frac{\sin4\alpha}{1+\cos4\alpha}\cdot\frac{\cos2\alpha}{1+\cos2\alpha}\cdot\frac{\cos\alpha}{1+\cos\alpha}=\tan\frac{\alpha}{2}.[/tex]
=
[tex]\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}.[/tex]
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:
[tex]\frac{\sin4\alpha}{1+\cos4\alpha}\cdot\frac{\cos2\alpha}{1+\cos2\alpha}\cdot\frac{\cos\alpha}{1+\cos\alpha}=\tan\frac{\alpha}{2}.[/tex]
=
[tex]\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}.[/tex]
I don't know how to use substitutions

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