# Thread: [SOLVED] More verifying that each trigonometric equation is an identity... again.

1. ## [SOLVED] More verifying that each trigonometric equation is an identity... again.

i still struggle with these;. i've been working on this one for a half hour.

(cot^2 x -1)/(cot^2 x +1)= 2sin^2 x

2. Originally Posted by somanyquestions
i still struggle with these;. i've been working on this one for a half hour.

(cot^2 x -1)/(cot^2 x +1)= 2sin^2 x
It's not an identity. The RHS should be 1 - 2 sin^2 x.

3. sorry about that. i worked on the corrected problem for a half hour.

4. Originally Posted by somanyquestions
sorry about that. i worked on the corrected problem for a half hour.
Substitute $\cot x = \frac{\cos x}{\sin x}$ into the LHS. Then multiply the resulting numerator and denominator by $\sin^2 x$. Then apply a well known double angle formula.

5. i never learned those formulas.. or maybe we're not there yet?? i don't know.

so..
(cot^2 x -1)/(cot^2 x +1)= 1-2sin^2 x

[(cos^2 x/ sin^2 x) -1] / [(cos^2 x/sin^2 x) +1]

[(cos^2x*sin^2 x/sin^4 x) -1]/[(cos^2x*sin^2 x/sin^4 x) +1]

?
then.. i don't know. i'll just ask my teacher, maybe she has some way to do it using methods i have already learned.

6. Originally Posted by somanyquestions
i never learned those formulas.. or maybe we're not there yet?? i don't know.

so..
(cot^2 x -1)/(cot^2 x +1)= 1-2sin^2 x

[(cos^2 x/ sin^2 x) -1] / [(cos^2 x/sin^2 x) +1]

Mr F says: When you multiply the numerator and denominator of the above line you get ${\color{red}\frac{\cos^2 x - \sin^2 x}{\cos^2 x + \sin^2 x}}$. I don't know how you got the line below.

[(cos^2x*sin^2 x/sin^4 x) -1]/[(cos^2x*sin^2 x/sin^4 x) +1]

?
then.. i don't know. i'll just ask my teacher, maybe she has some way to do it using methods i have already learned.
Are you saying you never learned that cot x = 1/tan x and that you don't know that tan x = sin x/cos x? And you're unfamiliar with the fact that $\cos^2 x = 1 - \sin^2 x$? (you don't actually need any double angle formula).

7. Originally Posted by mr fantastic
Are you saying you never learned that cot x = 1/tan x and that you don't know that tan x = sin x/cos x? And you're unfamiliar with the fact that $\cos^2 x = 1 - \sin^2 x$? (you don't actually need any double angle formula).
i am. i won't ask for help on these type of questions anymore. you make me feel unintelligent. this is why i need help. i can't even multiply this right.

8. Originally Posted by somanyquestions
i am. i won't ask for help on these type of questions anymore. you make me feel unintelligent. this is why i need help. i can't even multiply this right.
If you're given questions like the one you posted then you're expected to know the things I mentioned in my previous post. The application of these things in a particular question is what you can be helped with. But if you don't know these things, then you need to go back to your class notes and textbook and review that material so that you do know them.

9. Originally Posted by mr fantastic
If you're given questions like the one you posted then you're expected to know the things I mentioned in my previous post. The application of these things in a particular question is what you can be helped with. But if you don't know these things, then you need to go back to your class notes and textbook and review that material so that you do know them.
i know them except the angle rule or some other that people have been saying to me.. i don't know why i haven't learned that in class yet. i guess i just can't multiply them properly half the time. i'll work on it.

10. Originally Posted by mr fantastic
Are you saying you never learned that cot x = 1/tan x and that you don't know that tan x = sin x/cos x? And you're unfamiliar with the fact that $\cos^2 x = 1 - \sin^2 x$? (you don't actually need any double angle formula).