1. ## Establishing Trig Identities

Ive been working on this problem for quite some time now and i haven't been able to figure it out.

tanU(cscU-sinU)=cosU

this is what ive done so far

tanU*cscU - tanU*sinU

(sinU/cosU)*(1/sinU) - (sinU/cosU)*sinU

(1/cosU) - (sin^2U/cosU)

(1-sin^2(U)/cosU)

Half Angles Formula
(1-1+cos(2U))/2cosU

cos(2U)/2cosU

Double angles Formula
(2cos^2(U)-1)/2cosU

cosU-1

cosU-1 =/= cosU

ands so i end up with that pesky -1 out there.

so what am i doing wrong?

2. Originally Posted by skipmonkey
Ive been working on this problem for quite some time now and i haven't been able to figure it out.

tanU(cscU-sinU)=cosU

this is what ive done so far

tanU*cscU - tanU*sinU

(sinU/cosU)*(1/sinU) - (sinU/cosU)*sinU

(1/cosU) - (sin^2U/cosU)

(1-sin^2(U)/cosU)

Half Angles Formula
(1-1+cos(2U))/2cosU

cos(2U)/2cosU

Double angles Formula
(2cos^2(U)-1)/2cosU

cosU-1

cosU-1 =/= cosU

ands so i end up with that pesky -1 out there.

so what am i doing wrong?

$\frac{2\cos^2(U)-1}{2\cos(U)} \neq \cos(U)-1$

$\frac{2\cos^2(U)-1}{2\cos(U)} = \frac{\cos^2(U)-1}{\cos(U)}$

You could do two things now. You can split the fraction:

$= \frac{\cos^2(U)}{\cos(U)} - \frac{1}{\cos(U)} = \cos(U) - \sec(U)$

Or you could use the trigonometric identity $\sin^2x+\cos^2x=1$

$\frac{\cos^2(U)-1}{\cos(U)} = \frac{-\sin^2x}{\cos(U)} = -\sin{x} \tan{x}$

I don't know why you want the answer to be cos(U). Is that the answer in your book?

I think that last step should give you $\cos(U) - \sec(U)$ or $-\sin{x} \tan{x}$, which are both the same thing.

Hope that helps

Mathemagister

3. yes the cos(U) is part of the original question, we have to establish the identities by making both sides equal.

this is a test which two people in the entire class passed, so the teacher is letting us redo it at home for partial credit and i just cant figure this stuff out.

it was a 26 question test which we had an hour and a half to do. this is number 1. so I'll probably be back with more questions as i go through it