# Thread: Can't prove these identities?

1. ## Can't prove these identities?

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

From there I don't know where to go...

2) tan(pi/4 + x) + tan(pi/4 - x) = 2sec2x

For this one I don't even know where to start, I can't even find a formula that I could use for the LS.

All and any help is very much welcomed!

2. 1) sin(x+y)sin(x-y) = cos^2 y - cos^2 x
(sin x cos y + cos x sin y)(sin x cos y - cos x sin y) = cos^2 y - cos^2 x

From there I don't know where to go...
From there, use the difference of 2 squares.

Then, use the Pythagorean identity to substitute the sin by cos and you'll get it.

2. Start by using the compound angle identity for tan!

3. Originally Posted by doodofnood
1) sin(x+y)sin(x-y) = cos^2 y - cos^2 x
(sin x cos y + cos x sin y)(sin x cos y - cos x sin y) = cos^2 y - cos^2 x
$\displaystyle (a+b)(a-b) = a^2-b^2$ (difference of two squares).
2) tan(pi/4 + x) + tan(pi/4 - x) = 2sec2x
For this one I don't even know where to start, I can't even find a formula that I could use for the LS.
$\displaystyle \displaystyle \tan(x+y) = \frac{\tan{x}+\tan{y}}{1-\tan{x}\tan{y}}$

Oh, late. Ah, well...

4. so for the first 1 i got down to
sin^2xcos^2y - cos^2xsin^2y
= (1-cos^2x)cos^y - cos^2x(1-cos^2y)

now what...?

5. Originally Posted by doodofnood
now what...?
Well, why did you stop? Keep going, expand the little barackets.

6. so if i expand i get..

cos^2y - cos^4xy - cos^2x - cos^4xy
then cos^4xy cancels out?
:O

7. Originally Posted by doodofnood
so if i expand i get..

cos^2y - cos^4xy - cos^2x - cos^4xy
then cos^4xy cancels out?
:O
$\displaystyle (1-cos^2x)cos^2y - cos^2x(1-cos^2y)$
$\displaystyle = cos^2y -cos^4y-cos^2x+cos^4y$
now the $\displaystyle cos^4y$ cancel out and you are left with..