Express Trigonometric Identities of Terms of Variables

1. Dec 6, 2008

TheShehanigan

1. The problem statement, all variables and given/known data

Be x an element in the interval [Pi/4, 3Pi/4] express cos(2x), sin x, sin (x+pi) in terms of x. You must know that, for this question, cos x = z and z will always be < 0.

2. Relevant equations

cos(2x) = 2 (cos(x))^2 - 1
cos(2x) = cos^2 x - sin ^2 x
sin^2 x + cos ^2 x = 1
sin(x+pi) = -sin x

3. The attempt at a solution

I'm doing this for a friend, and it's been ages since I have tried this type of problem. Anyways, for cos(2x) I just used the identity cos(2x) = cos^2 x - sen ^2 x which can be proven with the Pythagorean theorem to be actually another form of cos(2x) = 2 cos^2 x - 1. I don't know, though, if that would be the final answer of if I need to express it further (since basically I have everything expressed as constants and cos), making what I think is the answer 2z^2 - 1.

I have really no idea for the sin(x), but I guess I will be using the Pythagorean identity. I got this document like 5 minutes ago and I've got to leave, so yeah. Just skimmed it and nothing came to mind.

The real problem is, though, sin(x+pi). If sin(x+pi) = -sin x using the sum of angles formula (and everything with a cosine cancels in there) how can I express sen(x+pi) in terms of the cosine?

Thanks for the help.

2. Dec 6, 2008

HallsofIvy

Staff Emeritus
It's not clear to me what you are asking. You say that the problem is "Be x an element in the interval [Pi/4, 3Pi/4] express cos(2x), sin x, sin (x+pi) in terms of x."

All of those are already in terms of x! Since you mention "You must know that, for this question, cos x = z " is it possible that you mean either "express cos(2x), sin x, sin (x+pi) in terms of z"?

If so then use the formulas you give:
cos(2x)= 2 (cos(x))2 - 1= 2z2- 1

$$sin(x)= \pm\sqrt{1- cos^2(x)}= \pm\sqrt{1- z^2}$$
but you will have to be careful about that sign: it can't stay "$\pm$"!

$$sin(x+\pi)= - sin(x)= -(\pm\sqrt{1- cos^2(x)})= -(\pm\sqrt{1- z^2})$$
and again you will have to think about what sign to put on this.

3. Dec 6, 2008

TheShehanigan

Right. My bad, I forgot to write that last part of the instructions.

Obviously, given the z<0 part of the answer and the domain of the function, then the first two shall be negative and the last one will be positive.

It was as I thought, so thank you for clearing that out. Sucks when you can't remember old stuff you did ages ago.