# Thread: trigonometric identity how should i start this

1. ## trigonometric identity how should i start this

sin theta/ sin theta - cos theta

= (convert to)

sin theta (sin theta - cos theta)/ 1 - 2 sin theta cos theta

how should i start this?
pls.. help

2. Originally Posted by xahjika
sin theta/ sin theta - cos theta

= (convert to)

sin theta (sin theta - cos theta)/ 1 - 2 sin theta cos theta

how should i start this?
pls.. help
Multiply $\displaystyle \frac{\sin{\theta}}{\sin{\theta}-\cos{\theta}}$ by $\displaystyle \frac{\sin{\theta}-\cos{\theta}}{\sin{\theta}-\cos{\theta}}$

This one should have been very obvious.

3. ok tnks.. haha..

by the way is this right?

= sin theta (sin theta - cos theta)/ (sin theta - cos theta)(sin theta - cos theta)

= sin theta (sin theta - cos theta)/ sin^2 theta - sin theta cos theta - cos theta sin theta + cos^2 theta

(sin^2 theta + cos^2 theta = 1)
(- sin theta cos theta - cos theta sin theta= -2 sin theta cos theta)

= sin theta (sin theta - cos theta)/ 1-2 sin theta cos theta

4. Originally Posted by xahjika
ok tnks.. haha..

by the way is this right?

= sin theta (sin theta - cos theta)/ (sin theta - cos theta)(sin theta - cos theta)

= sin theta (sin theta - cos theta)/ sin^2 theta - sin theta cos theta - cos theta sin theta + cos^2 theta

(sin^2 theta + cos^2 theta = 1)
(- sin theta cos theta - cos theta sin theta= -2 sin theta cos theta)

= sin theta (sin theta - cos theta)/ 1-2 sin theta cos theta
Bravo.

5. ## Re:

$\displaystyle \frac{sin\Theta}{sin\Theta-cos\Theta}=\frac{sin\Theta(sin\Theta-cos\Theta)}{1-2sin\Theta cos\Theta}$

I assume that you mean this .

From RHS ,
$\displaystyle \frac{sin\Theta(sin\Theta-cos\Theta)}{cos^2\Theta-2sin\Theta cos\Theta+sin^2\Theta}$
=$\displaystyle \frac{sin\Theta(sin\Theta-cos\Theta)}{(cos\Theta-sin\Theta)^2}$
=$\displaystyle \frac{-sin\Theta(cos\Theta-sin\Theta)}{(cos\Theta-sin\Theta)^2}$
=$\displaystyle \frac{-sin\Theta}{-(sin\Theta-cos\Theta)}$
=$\displaystyle \frac{sin\Theta}{sin\Theta-cos\Theta}$