# Proving if it's true (Equality)

• Dec 4th 2010, 08:07 PM
Cthul
Proving if it's true (Equality)
I'm no sure how to prove this equality true.
(Oh and I used an image instead of latex, because I don't know how, someone also tell me how to make to the power and cos?)
• Dec 4th 2010, 08:18 PM
Prove It
It doesn't. You need to use the identity

$\displaystyle \cos{(\alpha \pm \beta)} = \cos{\alpha}\cos{\beta} \mp \sin{\alpha}\sin{\beta}$.

So that means

$\displaystyle \cos{(x + y)}\cdot \cos{(x - y)} = (\cos{x}\cos{y} - \sin{x}\sin{y})(\cos{x}\cos{y} + \sin{x}\sin{y})$

$\displaystyle = \cos^2{x}\cos^2{y} - \sin^2{x}\sin^2{y}$

$\displaystyle = \cos^2{x}\cos^2{y} - (1 - \cos^2{x})(1 - \cos^2{y})$

$\displaystyle = \cos^2{x}\cos^2{y} - (1 - \cos^2{y} - \cos^2{x} + \cos^2{x}\cos^2{y})$

$\displaystyle = \cos^2{y} + \cos^2{x} - 1$.
• Dec 4th 2010, 08:21 PM
Cthul
Thanks, that was helpful.
(And I got how to do the latex now)
• Dec 5th 2010, 04:42 AM
When you get to

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

you can combine some of the squares to get 1.

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

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

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

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

$2cos^2y=cos^2x+sin^2x=1$

Note that there is a difference between an equation and an identity.
An identity is true for all x.
An equation is true for certain x for which we solve.
• Dec 5th 2010, 08:07 PM
Cthul
I have another that I'm not sure. I want to know how to prove its identity.
$\sin{\theta}\sin{2\theta}\cos{2\theta}=\frac{1}{4} (\cos{3\theta}-\cos{5\theta})$
• Dec 5th 2010, 08:18 PM
Prove It
Start by noting that $\displaystyle \cos{(5\theta)} = \cos{(3\theta + 2\theta)}$ and use the angle sum identity to simplify.

Then note that $\displaystyle \cos{(3\theta)} = \cos{(2\theta + \theta)}$ and again use the angle sum identity to simplify.
• Dec 6th 2010, 12:33 PM
Quote:

Originally Posted by Cthul
I have another that I'm not sure. I want to know how to prove its identity.
$\sin{\theta}\sin{2\theta}\cos{2\theta}=\frac{1}{4} (\cos{3\theta}-\cos{5\theta})$

Alternatively, get everything in terms of "Sine", using the following identities

$cosA-cosB=-2sin\left[\frac{A+B}{2}\right]sin\left[\frac{A-B}{2}\right]$

$sinAcosB=\frac{1}{2}\left[sin(A+B)+sin(A-B)\right]$

Then...

$sin\theta\;sin2\theta\;cos2\theta=sin\theta\left[\frac{1}{2}\left(sin4\theta+sin0\right)\right]=\frac{1}{2}sin\theta\;sin4\theta$

$\frac{1}{4}\left[cos3\theta-cos5\theta\right]=\frac{1}{4}\left[-2sin\left(\frac{8\theta}{2}\right)sin\left(-\frac{2\theta}{2}\right)\right]=-\frac{1}{2}\left[sin4\theta\left(-sin\theta\right)\right]$
• Dec 6th 2010, 03:05 PM
Soroban
Hello, Cthul!

Quote:

$\cos(x+y)\cos(x-y) \:=\:\cos^2\!x - \cos^2\!y$

I'm no sure how to prove this equality true.

I'm not surprised . . . The statement is not true.

• Dec 7th 2010, 10:59 AM
TheCoffeeMachine
Quote:
Why not just let $x = y = 0$, then it gives:

$\Rightarrow \cos\left(0\right)\cdot\cos\left(0\right) = \cos^2\left(0\right)-\cos^2\left(0\right)$

$\Rightarrow 1 = 0$. A counterexample!
• Dec 7th 2010, 12:13 PM
$y=\frac{\pi}{4};\;\;x=\frac{\pi}{4}$ is a solution among others...