http://209.85.48.8/9963/162/upload/p2222992.gif

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?)

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- Dec 4th 2010, 08:07 PMCthulProving if it's true (Equality)
http://209.85.48.8/9963/162/upload/p2222992.gif

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 PMProve It
It doesn't. You need to use the identity

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

So that means

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

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

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

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

$\displaystyle \displaystyle = \cos^2{y} + \cos^2{x} - 1$. - Dec 4th 2010, 08:21 PMCthul
Thanks, that was helpful.

(And I got how to do the latex now) - Dec 5th 2010, 04:42 AMArchie Meade
When you get to

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

you can combine some of the squares to get 1.

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

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

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

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

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

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 PMCthul
I have another that I'm not sure. I want to know how to prove its identity.

$\displaystyle \sin{\theta}\sin{2\theta}\cos{2\theta}=\frac{1}{4} (\cos{3\theta}-\cos{5\theta})$ - Dec 5th 2010, 08:18 PMProve It
Start by noting that $\displaystyle \displaystyle \cos{(5\theta)} = \cos{(3\theta + 2\theta)}$ and use the angle sum identity to simplify.

Then note that $\displaystyle \displaystyle \cos{(3\theta)} = \cos{(2\theta + \theta)}$ and again use the angle sum identity to simplify. - Dec 6th 2010, 12:33 PMArchie Meade
Alternatively, get everything in terms of "Sine", using the following identities

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

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

Then...

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

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

Quote:

$\displaystyle \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 AMTheCoffeeMachine
- Dec 7th 2010, 12:13 PMArchie Meade
Though if anyone got suckered into thinking it was an equation to be solved, then

$\displaystyle y=\frac{\pi}{4};\;\;x=\frac{\pi}{4}$ is a solution among others...

(Doh) - Dec 8th 2010, 01:26 AMCthul
Thanks for your response.