1. ## Pythagorean identities

Use Pythagorean identities to write the expression as an integer.

tan^2 4β - sec^2 4β

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

tan^2 4β - sec^2 4β

hint ...

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

3. lol...omg i'm still lost...

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

is this the next step, i dont understand how you get it to equal to an integer...i.e. get rid of the beta

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

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

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

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

6. Oh ok...wow i knew how to do it then...
so what if it was something like
$\displaystyle 4 tan^2(\beta)-4sec^2(\beta)$

7. 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?

8. 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

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

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

$\displaystyle 4( -1)$

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

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

10. What are you trying to do? Just simplify?

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

make $\displaystyle x =\cot\alpha$

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

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

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

$\displaystyle \frac{\cot\alpha+ 2}{\cot\alpha -3}$

11. yes simplifry the expression

12. Originally Posted by purplec16
yes simplifry the expression

Is done in the previous post.

Can also say

$\displaystyle \frac{\cot\alpha+ 2}{\cot\alpha -3} = 1+\frac{5}{\cot\alpha -3}$

13. 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. $\displaystyle 5(sin^2(\theta/4)+cos^2(\theta/4))$
$\displaystyle 5(1)$

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

Now factor out the 5 and follow what I did in post #8.

15. Ok, thank you so much, I did that

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