Difficult Trig Identity Question

**Diesal** · November 28th 2010, 07:02 PM

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?

**harish21** · November 28th 2010, 07:07 PM

Originally Posted by Diesal

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

Do you mean $\sin (x) = \sqrt{1+\cot^2 (x)}$ ?? this identity is not valid/true.

Originally Posted by Diesal

The identity looks like this:
-sinx = √1
- 1+cot^2x

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

What have you written is hard to understand. Please post your complete question clearly.

**Diesal** · November 28th 2010, 07:17 PM

How do I write it the way you did it?

**harish21** · November 28th 2010, 07:20 PM

Learn LaTeX. You can download the Latex.pdf tutorial and look at the basic commands.

**Diesal** · November 28th 2010, 07:27 PM

**harish21** · November 28th 2010, 07:34 PM

Use $\cot^2(x)=\dfrac{\cos^2(x)}{\sin^2(x)}$

and $\sin^2(x)+\cos^2(x)=1$

**Diesal** · November 28th 2010, 07:41 PM

How'd you solve it? I need to show my work, plus knowing how it was done will help me in the future.

**Educated** · November 28th 2010, 08:03 PM

$\sqrt{\dfrac{1}{1+\cot^2 (x)}}$

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

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

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

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

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

$=\sin (x)$

**Diesal** · November 28th 2010, 08:07 PM

Educated, you really are educated. Thank you.

**SammyS** · December 3rd 2010, 08:57 PM

Originally Posted by Diesal

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?

$\displaystyle \sin x = {1\over{\csc x}}$

$\displaystyle =\sqrt{1\over{\csc^2 x}}$

$\displaystyle =\sqrt{1\over{1+\cot^2 x}}$

**Soroban** · December 4th 2010, 06:07 AM

Hello, Diesal!

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

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

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