Simplifying Expression to tan $\frac{\alpha}{2}$

[ritwik06]

Homework Statement

Prove
\frac{sin 2 \alpha}{1+cos 2 \alpha}-\frac{cos \alpha}{1+cos \alpha}=tan \frac{ \alpha}{2}




Use half and double angle formulas, I get;

The given expression is equal to:
\frac{(2 tan \frac{\alpha}{2})(1+tan^{2} \frac{\alpha}{2})}{1-tan^{2}\frac{\alpha}{2}}+2 tan \frac{\alpha}{2}+tan^{2}\frac{\alpha}{2}-1

Please help me simplify this to only tan \frac{\alpha}{2}

 

[konthelion]

That form looks quite complicated.

Why don't you try expanding the RHS instead of the LHS using the half angle formula
tan \frac{\alpha}{2}= \pm \sqrt_{\frac{1-cos \alpha}{1+cos \alpha} = \frac{sin\alpha}{1+cos\alpha}

 

[ritwik06]

That form looks quite complicated.

Why don't you try expanding the RHS instead of the LHS using the half angle formula
tan \frac{\alpha}{2}= \pm \sqrt_{\frac{1-cos \alpha}{1+cos \alpha} = \frac{sin\alpha}{1+cos\alpha}

This implies that:
2 tan \alpha - \frac{cos \alpha}{1+ cos \alpha}=\frac{sin \alpha}{1+ cos \alpha} Isnt it?

But now the denominators are different. How will I proceed?

Thanks

 

[konthelion]

Oh darn. I didn't turn out like I hoped it would. Lol. Let's try your original method.

Let's see, by the double-angle formula,
tan(2\alpha)=\frac{2tan \alpha}{1-tan^2 \alpha} (*)

therefore, your original LHS

<br /> \frac{(2 tan \frac{\alpha}{2})(1+tan^{2} \frac{\alpha}{2})}{1-tan^{2}\frac{\alpha}{2}}+2 tan \frac{\alpha}{2}+tan^{2}\frac{\alpha}{2}-1<br />

then for the \frac{(2 tan \frac{\alpha}{2})(1+tan^{2} \frac{\alpha}{2})}{1-tan^{2}\frac{\alpha}{2}} term


By (*) 1-tan^2 \frac{\alpha}{2}= \frac{2tan \frac{\alpha}{2}}{tan \alpha}, so the \frac{(2 tan \frac{\alpha}{2}) should cancel

Edit:
This simplifies the LHS into
tan \alpha \left( 1+tan^2 \frac{\alpha}{2} \right) + +2 tan \frac{\alpha}{2}+tan^{2}\frac{\alpha}{2}-1

 

[Defennder]

Did you transcribe the question correctly? It appears that the "identity" doesn't hold for \alpha = \frac{\pi}{4}

 

[tiny-tim]

Hi ritwik06!

Homework Statement

Prove
\frac{sin 2 \alpha}{1+cos 2 \alpha}-\frac{cos \alpha}{1+cos \alpha}=tan \frac{ \alpha}{2}

ooh!

Golden rule: whnever you see (1 + cos) or (1 - cos), use the standard trigonometric identities for them!