# Thread: Help with trig identities

1. ## Help with trig identities

verify the identity: tanxcos(squared)x=[(2tanxcos(squared)x-tanx)/(1-tan(squared)x]

$\frac{sin(x)}{cos(x)}\cdot[cos(x)^2]=\frac{\frac{2sin(x)}{cos(x)^3}-\frac{sin(x)}{cos(x)}}{sec(x)^2-2}$..can you see whta to do from there

3. Originally Posted by kelsey3
verify the identity: tanxcos(squared)x=[(2tanxcos(squared)x-tanx)/(1-tan(squared)x]
with identities, always start with the more complicated side (the right side in this case). change everything to sine and cosines and then simplify

recommendations here: factor out the tangent first. remember your identities for cos(2x)

4. Originally Posted by Mathstud28
$\frac{sin(x)}{cos(x)}\cdot[cos(x)^2]=\frac{\frac{2sin(x)}{cos(x)^3}-\frac{sin(x)}{cos(x)}}{sec(x)^2-2}$..can you see whta to do from there
it's actually not that bad a problem if you factor out tangent first...

5. ## Haha

I think your secondary purpose on here is to help people with math...I think your primary purpose is to point out a highschooler(me)'s flaws..haha just kidding

6. Originally Posted by Mathstud28
I think your secondary purpose on here is to help people with math...I think your primary purpose is to point out a highschooler(me)'s flaws..haha just kidding
hehe, i was thinking the same thing, haha. it seems i have been picking on you a lot tonight, huh?

don't worry about it, when i just started here, the same thing happened to me. just look through some of my old posts, i was constantly making a fool of myself (i still do sometimes).

it's tough love. i have nothing against you. trust me, it will help you in the long run. if you feel like someone's always there to critique you, you'll start to think things through more carefully.

Power to you!

7. ya no kiddin....i dont really get how you got the sec(squared)x-2 on the bottom and where the cos(squared)x went. I don't know what identity you used.

8. ## Here you go

$sec(x)^2=tan(x)^2+1$ and I made a little error...dont worry I know what I am doing the cos(x)^2 went to the bottom

9. man im retarded......i got (sinx/cosx)/(1/cos(squared)x-2) and then i brought the bottom up and crossed out a cosine but its still not right

10. ## I am sorry

what exactly does that mean..show me your steps and I will critque you...use LaTeX if you can

11. how do i use that?....im new at this...

12. i just don't know how to get rid of the 2 on the bottom

13. ## To write it do this

the most important thing you will need to know is the fraction $\frac{cos(x)}{sin(x)}=cot(x)$ is made by[ math]\frac{cos(x)}{sin(x)}=cot(x)[/math but i took off the ] after /math...look up LaTeX if you plan on using this site often it is a neccessity to learn

14. Dude.....that looks way too complicated. I just dont know how to get rid of the 2 on the bottom....i get the rest

15. Originally Posted by kelsey3
how do i use that?....im new at this...
see the tutorial here for how to use LaTeX

i will start you off.

Consider the RHS:

$\displaystyle \frac {2 \tan x \cos^2 x - \tan x}{1 - \tan^2 x} = \tan x \cdot \frac {2 \cos^2 x - 1}{1 - \frac {\sin^2 x}{\cos^2 x}}$

$\displaystyle = \tan x \cdot \frac {2 \cos^2 x - 1}{\frac {\cos^2 x - \sin^2 x}{\cos^2 x}}$

$= \frac {\sin x}{\cos x} \cdot {\color{red}2 \cos^2 x - 1} \cdot \frac {\cos^2 x}{{\color{red}\cos^2 x - \sin^2 x}}$

now what? (of course, i put the two things in red for a reason)

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