# Thread: Analytic Trig- Sin, Cos, Tan

1. ## Analytic Trig- Sin, Cos, Tan

Hi, I was wondering how to solve this problem....

"Write each of the other trigonometric functins of x in tems of sin x."

I know that sin x= (1/csc x), but that's about it.

Thanks!

3. Originally Posted by checkthiskid

List of trigonometric identities - Wikipedia, the free encyclopedia
I appreciate your help.... I think it's supposed to be a picture? I'm sorry for some reason it's not working!

4. Originally Posted by falloutatthedisco
I appreciate your help.... I think it's supposed to be a picture? I'm sorry for some reason it's not working!
yeah im rubbish with computers

List of trigonometric identities - Wikipedia, the free encyclopedia

it was supposed to be a link but i still had the image copied and pasted
scroll down and there will be a table of all the identities

5. For example,

$\tan x=\frac{\sin x}{\cos x}=\frac{\sin x}{\sqrt{1-{{\sin }^{2}}x}}\implies {{\tan }^{2}}x=\frac{{{\sin }^{2}}x}{1-{{\sin }^{2}}x},$ so $\sin x=\frac{\tan x}{\sqrt{1+{{\tan }^{2}}x}}.$

Obviously, this requires restrictions for $x.$ For example, $\cos x=\sqrt{1-\sin^2x}$ works when $0\le x\le\frac\pi2$ or $\frac{3\pi}2\le x\le2\pi.$

6. Checkthiskid-

it's fine.... I'm rubbish at pre-calc!

thanks so much for the help!

Krisalid-

thanks so much!

7. Originally Posted by Krizalid
For example,

$\tan x=\frac{\sin x}{\cos x}=\frac{\sin x}{\sqrt{1-{{\sin }^{2}}x}}\implies {{\tan }^{2}}x=\frac{{{\sin }^{2}}x}{1-{{\sin }^{2}}x},$ so $\sin x=\frac{\tan x}{\sqrt{1+{{\tan }^{2}}x}}.$

Obviously, this requires restrictions for $x.$ For example, $\cos x=\sqrt{1-\sin^2x}$ works when $0\le x\le\frac\pi2$ or $\frac{3\pi}2\le x\le2\pi.$
you may need to show your working out so i would suggest doing it step by step using what was shown in krizalid's post and using $cosx=\sqrt{1-sin^{2}x}$
derived from $cos^{2}x +sin^{2}x=1$