# Simple triganomic identity

1. Mar 22, 2005

### lektor

Nothing special, i'm just not clicking on this question

$$\frac {1-\cos\theta}{\sin\theta} = \tan \frac {a}{2}$$

cheers

2. Mar 22, 2005

### Data

I'm going to assume by $a$ you mean $\theta$.

In that case, use

$$\cos^2{u} = \frac{1}{2}(1+ \cos{2u})$$

and

$$\sin^2{u} = \frac{1}{2}(1 - \cos{2u})$$

$$\Longrightarrow \tan^2{u} = \frac{1 - \cos{2u}}{1 + \cos{2u}}$$

to get rid of the annoying $$\frac{\theta}{2}$$, and then multiply by an appropriate version of 1.

3. Mar 22, 2005

### dextercioby

$$\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$$

$$1-\cos\theta=2\sin^{2}\frac{\theta}{2}$$

Daniel.

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