1. Apr 17, 2007

### r_maths

http://img117.imageshack.us/img117/1482/graph015tt6.png [Broken]
I know an alternative way of showing Sin C is cos 'theta'.
Thanks

Last edited by a moderator: May 2, 2017
2. Apr 17, 2007

### Feldoh

What is angle C? I can't make it out in the picture... Also could you show your attempt of the problems?

3. Apr 17, 2007

### r_maths

Angle C is pi/2 - theta (radians) "pi over two minus theta" ie. 90 deegrees - theta.
I did show my attempt at it, i couldn't get further.

4. Apr 18, 2007

### HallsofIvy

Staff Emeritus
Certainly, since the angles in a triangle add to $2\pi$ radians, $B= 2\pi- \lambda- (\pi/2- \theta)= 3\pi/2- (\theta- \lambda)$.
Okay, using sin(a+ b)= sin(a)cos(b)- cos(a)sin(b)[/itex], what is
$sin(3\pi/2+ (-(\theta- \lambda)))$?

5. Apr 18, 2007

### cristo

Staff Emeritus
The angles of a triangle add upto $\pi$ radians
thus $B=\pi-\lambda-(\pi/2-\theta)=\pi/2+(\theta-\lambda)$
The expression you require is therefore $\sin(\pi/2+(\theta-\lambda))$ = ?

6. Apr 18, 2007

### r_maths

$\cos(\theta-\lambda)$

Thanks