Express Trigonometric Identities of Terms of Variables

[TheShehanigan]

Homework Statement

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.

Homework 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

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.

 

[HallsofIvy]

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.

 

[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.