Okay, for this one I need to somehow get sinx to equal to the square root of 1 over 1+cot(squared)x.

The identity looks like this:

-sinx = √1

-------1+cot^2x

So could someone show me in detail how I would solve this identity?

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- Nov 28th 2010, 07:02 PMDiesalDifficult Trig Identity Question
Okay, for this one I need to somehow get sinx to equal to the square root of 1 over 1+cot(squared)x.

The identity looks like this:

-sinx = √__1__

-------1+cot^2x

So could someone show me in detail how I would solve this identity? - Nov 28th 2010, 07:07 PMharish21
- Nov 28th 2010, 07:17 PMDiesal
How do I write it the way you did it?

- Nov 28th 2010, 07:20 PMharish21
Learn LaTeX. You can download the Latex.pdf tutorial and look at the basic commands.

- Nov 28th 2010, 07:27 PMDiesal
- Nov 28th 2010, 07:34 PMharish21
Use $\displaystyle \cot^2(x)=\dfrac{\cos^2(x)}{\sin^2(x)}$

and $\displaystyle \sin^2(x)+\cos^2(x)=1$ - Nov 28th 2010, 07:41 PMDiesal
How'd you solve it? I need to show my work, plus knowing how it was done will help me in the future.

- Nov 28th 2010, 08:03 PMEducated
$\displaystyle \sqrt{\dfrac{1}{1+\cot^2 (x)}}$

$\displaystyle =\sqrt{\dfrac{1}{1+\frac{\cos^2 (x)}{\sin^2 (x)}}}$

$\displaystyle =\sqrt{\dfrac{1}{\frac{\sin^2 (x)}{\sin^2 (x)}+\frac{\cos^2 (x)}{\sin^2 (x)}}}$

$\displaystyle =\sqrt{\dfrac{1}{\frac{\cos^2 (x) + sin^2 (x)}{\sin^2 (x)}}}$

$\displaystyle =\sqrt{\dfrac{1}{\frac{1}{\sin^2 (x)}}}$

$\displaystyle =\sqrt{\sin^2 (x)}$

$\displaystyle =\sin (x)$ - Nov 28th 2010, 08:07 PMDiesal
Educated, you really are educated. Thank you.

- Dec 3rd 2010, 08:57 PMSammyS
- Dec 4th 2010, 06:07 AMSoroban
Hello, Diesal!

You probably know that: .$\displaystyle \sin^2\!x + \cos^2\!x \:=\:1$

But are you familiar with these? .$\displaystyle \begin{Bmatrix}1 + \tan^2\!x \;=\; \sec^2\!x \\ 1 + \cot^2\!x \;=\; \csc^2\!x\end{Bmatrix}$

Quote:

$\displaystyle \sin x \:=\:\sqrt{\dfrac{1}{1 + \cot^2\!x}} $

The right side is:

. . $\displaystyle \sqrt{\dfrac{1}{1 + \cot^2\!x}} \;=\;\sqrt{\dfrac{1}{\csc^2\!x}} \;=\;\sqrt{\sin^2\!x} \;=\;\sin x $

- Dec 5th 2010, 07:11 AMPunch
You should end off with $\displaystyle \sqrt{sin^x}=sinx$ OR $\displaystyle -sinx$

and remember to reject the negative answer