# Pythagorean identities

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• Jan 26th 2010, 05:12 PM
purplec16
Pythagorean identities
Use Pythagorean identities to write the expression as an integer.

tan^2 4β - sec^2 4β

• Jan 26th 2010, 05:58 PM
skeeter
Quote:

Originally Posted by purplec16
Use Pythagorean identities to write the expression as an integer.

tan^2 4β - sec^2 4β

hint ...

$\displaystyle 1 + \tan^2(4\beta) = \sec^2(4\beta)$
• Jan 26th 2010, 06:16 PM
purplec16
lol...omg i'm still lost...
• Jan 26th 2010, 06:25 PM
purplec16
$\displaystyle 1+tan^2(4\beta)-sec^2(4\beta)$
$\displaystyle tan^2(4\beta)-sec^2(4\beta)=-1$

is this the next step, i dont understand how you get it to equal to an integer...i.e. get rid of the beta
• Jan 26th 2010, 06:27 PM
pickslides
Skeeter has given you the answer. You don't need to get rid of beta.

$\displaystyle 1 + \tan^2(4\beta) = \sec^2(4\beta)$

$\displaystyle 1 + \tan^2(4\beta) - \sec^2(4\beta)=0$

$\displaystyle \tan^2(4\beta) - \sec^2(4\beta)=-1$
• Jan 26th 2010, 06:29 PM
purplec16
Oh ok...wow i knew how to do it then...
so what if it was something like
$\displaystyle 4 tan^2(\beta)-4sec^2(\beta)$
• Jan 26th 2010, 06:30 PM
purplec16
Oh ok...wow i knew how to do it then...
so what if it was something like
$\displaystyle 4 tan^2 (\beta)-4 sec^2 (\beta)$
what would happen in that case?
• Jan 26th 2010, 06:33 PM
pickslides
Quote:

Originally Posted by purplec16
Use Pythagorean identities to write the expression as an integer.

$\displaystyle \tan^2(4\beta) - \sec^2(4\beta)=-1$

$\displaystyle -1$ is an integer so you are done.

For the next question

$\displaystyle 4 tan^2 (\beta)-4 sec^2 (\beta)$

$\displaystyle 4( tan^2 (\beta)- sec^2 (\beta))$

$\displaystyle 4( -1)$
• Jan 26th 2010, 06:40 PM
purplec16
Sorry to bother you but would u be able to assist me in solving something like this:

$\displaystyle \frac{cot^2\alpha-4}{cot^2\alpha-cot\alpha-6}$
• Jan 26th 2010, 06:46 PM
pickslides
What are you trying to do? Just simplify?

$\displaystyle \frac{\cot^2\alpha-4}{\cot^2\alpha-\cot\alpha-6}$

make $\displaystyle x =\cot\alpha$

$\displaystyle \frac{x^2-4}{x^2-x-6}$

$\displaystyle \frac{(x-2)(x+2)}{(x-3)(x-2)}$

$\displaystyle \frac{x+2}{x-3}$

$\displaystyle \frac{\cot\alpha+ 2}{\cot\alpha -3}$
• Jan 26th 2010, 06:47 PM
purplec16
yes simplifry the expression
• Jan 26th 2010, 06:51 PM
pickslides
Quote:

Originally Posted by purplec16
yes simplifry the expression

Is done in the previous post.

Can also say

$\displaystyle \frac{\cot\alpha+ 2}{\cot\alpha -3} = 1+\frac{5}{\cot\alpha -3}$
• Jan 26th 2010, 07:16 PM
purplec16
Quick Question for this expression:

$\displaystyle 5sin^2(\theta/4)+5cos^2(\theta/4)$

how do u get rid of the $\displaystyle (\theta/4)$ to make it equal to 5? i understand that it will be equal to 5
is it that i dont have to worry about the theta and jus simplify it equal to five
i.e. $\displaystyle 5(sin^2(\theta/4)+cos^2(\theta/4))$
$\displaystyle 5(1)$
• Jan 26th 2010, 07:29 PM
pickslides
$\displaystyle \sin^2\left(\frac{\theta}{4}\right)+\cos^2\left(\f rac{\theta}{4}\right) = 1$

Now factor out the 5 and follow what I did in post #8.
• Jan 26th 2010, 07:31 PM
purplec16
Ok, thank you so much, I did that
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