# Trigonometry I

1. Jun 22, 2008

### ritwik06

1. The problem statement, all variables and given/known data

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 $$tan \frac{\alpha}{2}$$

Last edited: Jun 22, 2008
2. Jun 22, 2008

### konthelion

That form looks quite complicated.

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

Last edited: Jun 22, 2008
3. Jun 22, 2008

### ritwik06

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

4. Jun 22, 2008

### 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,
$$tan(2\alpha)=\frac{2tan \alpha}{1-tan^2 \alpha}$$ (*)

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
$$tan \alpha \left( 1+tan^2 \frac{\alpha}{2} \right) + +2 tan \frac{\alpha}{2}+tan^{2}\frac{\alpha}{2}-1$$

Last edited: Jun 22, 2008
5. Jun 22, 2008

### Defennder

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

6. Jun 23, 2008

### tiny-tim

Hi ritwik06!
ooh!

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