Pythagorean identities

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January 26th 2010, 06:12 PM
purplec16

Pythagorean identities

Use Pythagorean identities to write the expression as an integer.

tan^2 4β - sec^2 4β

Please help me i dont know what to do...
January 26th 2010, 06:58 PM
skeeter

Quote:

Originally Posted by purplec16

Use Pythagorean identities to write the expression as an integer.

tan^2 4β - sec^2 4β

Please help me i dont know what to do...

hint ...

$1 + \tan^2(4\beta) = \sec^2(4\beta)$
January 26th 2010, 07:16 PM
purplec16

lol...omg i'm still lost...
January 26th 2010, 07:25 PM
purplec16

$1+tan^2(4\beta)-sec^2(4\beta)<br />$
$tan^2(4\beta)-sec^2(4\beta)=-1<br />$

is this the next step, i dont understand how you get it to equal to an integer...i.e. get rid of the beta
January 26th 2010, 07:27 PM
pickslides

Skeeter has given you the answer. You don't need to get rid of beta.

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

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

$\tan^2(4\beta) - \sec^2(4\beta)=-1$
January 26th 2010, 07:29 PM
purplec16

Oh ok...wow i knew how to do it then...
so what if it was something like
$4 tan^2(\beta)-4sec^2(\beta)$
January 26th 2010, 07:30 PM
purplec16

Oh ok...wow i knew how to do it then...
so what if it was something like
$4 tan^2 (\beta)-4 sec^2 (\beta)$
what would happen in that case?
January 26th 2010, 07:33 PM
pickslides

Quote:

Originally Posted by purplec16

Use Pythagorean identities to write the expression as an integer.

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

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

For the next question

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

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

$4( -1)$
January 26th 2010, 07:40 PM
purplec16

Sorry to bother you but would u be able to assist me in solving something like this:

$\frac{cot^2\alpha-4}{cot^2\alpha-cot\alpha-6}$
January 26th 2010, 07:46 PM
pickslides

What are you trying to do? Just simplify?

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

make $x =\cot\alpha$

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

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

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

$\frac{\cot\alpha+ 2}{\cot\alpha -3}$
January 26th 2010, 07:47 PM
purplec16

yes simplifry the expression
January 26th 2010, 07:51 PM
pickslides

Quote:

Originally Posted by purplec16

yes simplifry the expression

Is done in the previous post.

Can also say

$\frac{\cot\alpha+ 2}{\cot\alpha -3} = 1+\frac{5}{\cot\alpha -3}$
January 26th 2010, 08:16 PM
purplec16

Quick Question for this expression:

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

how do u get rid of the $(\theta/4)$ 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. $5(sin^2(\theta/4)+cos^2(\theta/4))$
$5(1)$
January 26th 2010, 08:29 PM
pickslides

$\sin^2\left(\frac{\theta}{4}\right)+\cos^2\left(\f rac{\theta}{4}\right) = 1$

Now factor out the 5 and follow what I did in post #8.
January 26th 2010, 08:31 PM
purplec16

Ok, thank you so much, I did that

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