# Trigonometry assistance

1. Aug 18, 2005

### TonyC

How do I go about finding the equivalent expression for cot 50 degrees?

When I work the problem, I come up with tan 130 degrees.

Where am I going wrong?

2. Aug 18, 2005

### TD

I rather work in radians, if you don't mind.

$$50^\circ = \frac{{5\pi }}{{18}}$$

We know that:
$$\cot \left( \alpha \right) = \tan \left( {\frac{\pi }{2} - \alpha } \right)$$

So:
$$\cot \left( {\frac{{5\pi }}{{18}}} \right) = \tan \left( {\frac{\pi }{2} - \frac{{5\pi }}{{18}}} \right) = \tan \left( {\frac{{2\pi }}{9}} \right) = \tan \left( {40^\circ } \right)$$

3. Aug 18, 2005

### TonyC

That makes perfect sense now.....

The light has come on, thank you

4. Aug 18, 2005

No problem