Can't prove these identities?

• Nov 20th 2010, 06:16 AM
doodofnood
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!
• Nov 20th 2010, 06:24 AM
Unknown008
Quote:

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. (Smile)

2. Start by using the compound angle identity for tan!
• Nov 20th 2010, 06:33 AM
TheCoffeeMachine
Quote:

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

$(a+b)(a-b) = a^2-b^2$ (difference of two squares).
Quote:

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.
$\displaystyle \tan(x+y) = \frac{\tan{x}+\tan{y}}{1-\tan{x}\tan{y}}$

Oh, late. Ah, well... :)
• Nov 20th 2010, 06:46 AM
doodofnood
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...?
• Nov 20th 2010, 06:51 AM
TheCoffeeMachine
Quote:

Originally Posted by doodofnood
now what...?

Well, why did you stop? Keep going, expand the little barackets.
• Nov 20th 2010, 06:58 AM
doodofnood
so if i expand i get..

cos^2y - cos^4xy - cos^2x - cos^4xy
then cos^4xy cancels out?
:O
• Nov 20th 2010, 07:07 AM
harish21
Quote:

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

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

$(1-cos^2x)cos^2y - cos^2x(1-cos^2y)$
$= cos^2y -cos^4y-cos^2x+cos^4y$
now the $cos^4y$ cancel out and you are left with..