Trig Identities: cos^2(2x) + sin^2(2x), os^2(x)-1/sin(x),

[Lokito]

I've done a few problems on a homework assignment related to Trig identities, but I can't figure out how to simplify some problems. I have the identities in front of me, but I can't see how they can be applied. I missed class when we went over them.

4. cos^2(2x) + sin^2(2x)
(I don't know if cos^2 is the correct way to represent cosine squared, but that's what I mean, in any case)

Unless I'm making a big mistake, this is the same as
cos^2(x) + sin^2(x), which = 1 So, cos^2(2x) +sin^2(2x) = 1?

6. cos^2(x)-1/sin(x)
This could become
1-sin^2(x)-1/sin(x)
Which is -sin^2(x)/sin(x), or -sin(x). I think.

15. Use algebra to prove the identity:
sin(x)/1-cos(x) = 1+cos(x)/sin(x)

I *understand* it, but I don't know how to prove it. Even though it's an odd number, the book conveniently leaves out an explanation/answer in the back index.

Lastly, Question 29. An image-based question:
http://img255.imageshack.us/img255/962/question29xc9.jpg

The back of the book lists the answer to part a) as tan(theta). I don't understand that at all. Y isn't tan(theta), it's just the opposite side. How'd that happen?
Part b) is sin(theta), since cos(phi) = adjacent/hypotenuse, where the side adjacent to angle phi is opposite angle theta.
Part c: 1 + y^2 is... ? Somehow it's (1/cos^2(theta)), or tan^2(theta). I don't see how that works.
Part d is 1/2tan(theta)??? Tangent is sine over cosine, not sine times cosine.

Eh.

 

[tkhunny]

4. cos^2(2x) + sin^2(2x)
(I don't know if cos^2 is the correct way to represent cosine squared, but that's what I mean, in any case)

Unless I'm making a big mistake, this is the same as
cos^2(x) + sin^2(x), which = 1 So, cos^2(2x) +sin^2(2x) = 1?

Perfect. Never doubt again.

6. cos^2(x)-1/sin(x)
This could become
1-sin^2(x)-1/sin(x)
Which is -sin^2(x)/sin(x), or -sin(x). I think.

Too bad your notation is not clear enough. If you mean for the entire expression to be in the numerator, you will need to write it like this:
[cos^2(x)-1]/sin(x) = [1-sin^2(x)-1]/sin(x) = -sin^2(x)/sin(x) = -sin(x)

And, really, what do you have against the equal sign?

Also, why did you do this? cos^2(x)-1 = 1-sin^2(x)-1 That's a lot of effort fir a direct substitution to the end result.
1 - cos^2(x) = sin^2(x) AND cos^2(x) - 1 = -sin^2(x)

15. Use algebra to prove the identity:
sin(x)/1-cos(x) = 1+cos(x)/sin(x)

Again, notation is bad. Use parentheses to calrify meaning.

Think about "Difference of Squares" and apply that idea to the left-hand side of 1 - cos^2(x) = sin^2(x)