Help with trig identities

**kelsey3** · April 6th 2008, 07:40 PM

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

**Mathstud28** · April 6th 2008, 07:46 PM

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

**Jhevon** · April 6th 2008, 07:49 PM

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)

**Jhevon** · April 6th 2008, 07:50 PM

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...

**Mathstud28** · April 6th 2008, 07:51 PM

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

**Jhevon** · April 6th 2008, 07:55 PM

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!

**kelsey3** · April 6th 2008, 07:58 PM

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.

**Mathstud28** · April 6th 2008, 08:03 PM

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

**kelsey3** · April 6th 2008, 08:13 PM

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

**Mathstud28** · April 6th 2008, 08:16 PM

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

**kelsey3** · April 6th 2008, 08:18 PM

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

**kelsey3** · April 6th 2008, 08:19 PM

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

**Mathstud28** · April 6th 2008, 08:22 PM

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

**kelsey3** · April 6th 2008, 08:24 PM

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

**Jhevon** · April 6th 2008, 08:25 PM

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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Help with trig identities

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