1. ## Trig identity

I have partially solved a question.

Prove that:

(sin x)^2 - (sin y)^2 = (sin x cos y)^2 - (cos x sin y)^2

2. Originally Posted by Stuck Man
I have partially solved a question.

Prove that:

(sin x)^2 - (sin y)^2 = (sin x cos y)^2 - (cos x sin y)^2
$(\sin x)^2 - (\sin y)^2 = (\sin x \cos y)^2 - (\cos x \sin y)^2$

$\sin^2 x - \sin^2 y = \sin^2 x \cos^2 y - \cos^2 x \sin^2 y$

$\sin^2 x - \sin^2 y = (1-cos^2 x)\cos^2 y - \cos^2 x (1-\cos^2 y)$

$\sin^2 x - \sin^2 y = \cos^2 y-cos^2 x\cos^2 y - \cos^2 x + \cos^2 x\cos^2 y$

$\sin^2 x - \sin^2 y = \cos^2 y - \cos^2 x$

$\sin^2 x - \sin^2 y = 1-\sin^2 y -(1- \sin^2 x)$

$\sin^2 x - \sin^2 y = 1-\sin^2 y -1+\sin^2 x$

$\sin^2 x - \sin^2 y = \sin^2 x-\sin^2 y$

3. Thanks. I didn't know the formula used twice by the first line. Is there a standard formula for that?

4. Originally Posted by Stuck Man
Thanks. I didn't know the formula used twice by the first line. Is there a standard formula for that?
$\sin^2 x + \cos^2 x = 1$

5. I don't know what that has to do with it. I meant is the a formula for (sin x cos y)^2 = (sin x)^2 (cos y)^2?

6. Originally Posted by Stuck Man
I don't know what that has to do with it. I meant is the a formula for (sin x cos y)^2 = (sin x)^2 (cos y)^2?
It is not a formula.
Its one of the exponents' laws.
$(ab)^c=a^c \;\ b^c$

7. Thanks.