Thread: 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!
Thanks in advance

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!

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

(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.

Oh, late. Ah, well...

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...?

Originally Posted by doodofnood

now what...?

Well, why did you stop? Keep going, expand the little barackets.

so if i expand i get..

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

Originally Posted by doodofnood

so if i expand i get..

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

you made a small mistake:





now the cancel out and you are left with..