# TRIG Identities and Equations

• Mar 29th 2012, 08:13 PM
Crysland
TRIG Identities and Equations
These are the problems I have trouble with while doing a practice on this topic.

1. Factor the expression and use fundamental identities to simplify:
cos^(2)x csc^(2)x - cos^(2)x

2. Find all solutions in the interval [0 , 2pi)
tan^(2)θ = (-3/2)secθ

3. Find an expression that completes the identity
[2sin^(2)x + cos(2x)] / csc(x) =

It would be helpful if u could explain step by step. I'm not just looking for answers. Thanks
• Mar 29th 2012, 08:38 PM
pickslides
Re: TRIG Identities and Equations
In the first one, take out $\displaystyle \cos^2x$ as a factor, what do you get?

For the second, create a quadratic using $\displaystyle \tan^2\theta = \sec^2\theta -1$ what do you get?
• Mar 29th 2012, 10:28 PM
highvoltage
Re: TRIG Identities and Equations
Here is #2.
Attachment 23467
• Mar 29th 2012, 10:40 PM
biffboy
Re: TRIG Identities and Equations
In 3 use the identity cos2x=1-2sin^(2)x
• Mar 30th 2012, 06:54 AM
Soroban
Re: TRIG Identities and Equations
Hello, Crysland!

Quote:

$\displaystyle \text{3. Find an expression that completes the identity,}$

. . $\displaystyle \frac{2\sin^2\!x + \cos2x}{\csc x}\:=\:\_\_\_$

Replace $\displaystyle \cos2x$ with $\displaystyle \cos^2\!x - \sin^2\!x$

$\displaystyle \frac{2\sin^2\!x + \cos2x}{\csc x} \;=\;\frac{2\sin^2\!x + (\cos^2\!x - \sin^2\!x)}{\csc x}$

. . . . . . . . . . . $\displaystyle =\; \frac{\overbrace{\sin^2\!x + \cos^2\!x}^{\text{This is 1}}}{\csc x}$

. . . . . . . . . . . $\displaystyle =\; \frac{1}{\csc x}$

. . . . . . . . . . . $\displaystyle =\;\sin x$
• Apr 3rd 2012, 12:37 PM
Crysland
Re: TRIG Identities and Equations
thanks so much. I'm still stuck with #1.
I got 2 answers, cos^2x*cot^2x and cos^4x/sin^2x
but don't know which one is correct and simplified.
• Apr 3rd 2012, 11:25 PM
biffboy
Re: TRIG Identities and Equations