# Proving Trig Identity: Sine Difference

• Aug 30th 2009, 04:17 PM
lucius
Proving Trig Identity: Sine Difference
In this figure, the triangle has hypotenuse 1.

I need to use it to show that:

$\displaystyle sin(A - B) = sinAcosB - sinBcosA$

The hint is :
$\displaystyle sin(A - B) = \frac{X}{h}$
Find other expressions for X and h and substitute them into this equation.

I know this is easy. But I'm just stuck. I've tried working out other side lengths and looking at other trig identities but I'm just at a brick wall without a clue why.

Thanks for any help.
• Aug 30th 2009, 09:19 PM
I-Think
Greetings. You don't need any other trig identities. Everything is on the diagram

Step 1. Fill in the angles, and you'll see that the 3rd from left triangle and the large triangle are similar.

Step 2. Find the scale factor.
In this case, the large triangle has a hypotenuse of 1 and the small triangle has a hypotenuse of $\displaystyle sinA-hsinB$
Hence scale factor(Q): $\displaystyle Q=\frac{1}{sinA-hsinB}$

Step 3. Apply this to the other side of the triangle, opposite the angle of $\displaystyle (90-A)$.
$\displaystyle QX=cosA$
Hence: $\displaystyle X=cosAsinA-hcosAsinB$

Put into http://www.mathhelpforum.com/math-he...8ed8cd8a-1.gif
And you get: $\displaystyle sin(A-B)=\frac{cosAsinA-hcosAsinB}{h}$

Simplify, then look at your triangle and realize that
$\displaystyle cosB=\frac{cosA}{h}$

Substitute and you arrive at the formula
http://www.mathhelpforum.com/math-he...e03d10b0-1.gif
• Sep 4th 2009, 04:02 AM
lucius
Excellent solution I-think!

Cleared that right up.

Thank you so much.