# identities

• Apr 28th 2008, 03:09 PM
identities
identities
• Apr 28th 2008, 03:14 PM
Moo
Hello,

Well, do you know identities for cos(a+b) and cos(a-b) ?

cos(a+b)=cos(a)cos(b)-sin(a)sin(b)

So what will cos(a-b) be ?

For the second one, use twice the formula for cos(a+b) :

cos(3x)=cos(2x+x)=...
• Apr 28th 2008, 03:22 PM
o_O
$\cos(A + B) = {\color{red}\cos A \cos B} - \sin A \sin B$
$\cos(A - B) = {\color{red} \cos A \cos B} + \sin A \sin B$

If A = B:
$\cos (2A) = \cos(A + A) = \cos A \cos A - \sin A \sin A = \cos^{2} A - \sin^{2} A$

Convert into terms of only cosA and you should get your answer

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Use your result from the first question to split your products in to sums. It'll be easier to integrate.

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$x = \cos t \: \: \Rightarrow \: \: dx = -\sin t dt$

Make the subs:

$\int \frac{\cos^{2} t}{\sqrt{1 - \cos^{2} t}} \: \cdot -\sin t dt \: \: = \: \:- \int \frac{\cos^{2} t \sin t}{\sqrt{\sin^{2} t}} dt$

etc.

You'll need your half-angle identity from your first question to do it after you simplify.
• Apr 28th 2008, 03:26 PM
Mathstud28
Quote:

identities

For the third part $\int\frac{x^2}{\sqrt{1-x^2}}$

let $x=\cos(\theta)$...dont often see this sub...
then $dx=-\sin(\theta)$
and $x^2=\cos^2(\theta)$

we need to find the limits of integration so $\frac{1}{2}=\cos(\theta)\Rightarrow(\theta=\frac{\ pi}{3}$

and $0=\cos(\theta)\Rightarrow{\theta=\frac{\pi}{2}}$

So putting that in we have

$\int_{\frac{\pi}{2}}^{\frac{\pi}{3}}\frac{\cos^2(\ theta)\cdot{-\sin(\theta)}d\theta}{\sqrt{1-\cos^2(\theta)}}$ $=\int_{\frac{\pi}{2}}^{\frac{\pi}{3}}\frac{\cos^2( \theta)-\sin(\theta)d\theta}{\sin(\theta)}$

Simplifying we get $-\int_{\frac{\pi}{2}}^{\frac{\pi}{3}}\cos^2(\theta) d\theta$

Using the identity $\cos^2(\theta)=\frac{1+\cos(2\theta)}{2}$

we see this is equal to $-\bigg[\frac{1}{2}\sin(2\theta)+\frac{\theta}{2}\bigg|_{\ frac{\pi}{2}}^{\frac{\pi}{3}}=.045293$
• Apr 28th 2008, 03:28 PM
Mathstud28
Quote:

Originally Posted by o_O
$\cos(A + B) = {\color{red}\cos A \cos B} - \sin A \sin B$
$\cos(A - B) = {\color{red} \cos A \cos B} + \sin A \sin B$

If A = B:
$\cos (2A) = \cos(A + A) = \cos A \cos A - \sin A \sin A = \cos^{2} A - \sin^{2} A$

Convert into terms of only cosA and you should get your answer

---------

Use your result from the first question to split your products in to sums. It'll be easier to integrate.

---------

$x = \cos t \: \: \Rightarrow \: \: dx = -\sin t dt$

Make the subs:

$\int \frac{\cos^{2} t}{\sqrt{1 - \cos^{2} t}} \: \cdot -\sin t dt \: \: = \: \:- \int \frac{\cos^{2} t \sin t}{\sqrt{\sin^{2} t}} dt$

etc.

You'll need your half-angle identity from your first question to do it after you simplify.

Darn you o_O! Well...its 1-2..but I think it should be 1.5-2 since I put the details (Crying)....:D