# Difficult Trig Identity Question

• Nov 28th 2010, 07:02 PM
Diesal
Difficult 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 PM
harish21
Quote:

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.

Quote:

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.
• Nov 28th 2010, 07:17 PM
Diesal
How do I write it the way you did it?
• Nov 28th 2010, 07:20 PM
harish21
Learn LaTeX. You can download the Latex.pdf tutorial and look at the basic commands.
• Nov 28th 2010, 07:27 PM
Diesal
• Nov 28th 2010, 07:34 PM
harish21
Use $\cot^2(x)=\dfrac{\cos^2(x)}{\sin^2(x)}$

and $\sin^2(x)+\cos^2(x)=1$
• Nov 28th 2010, 07:41 PM
Diesal
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 PM
Educated
$\sqrt{\dfrac{1}{1+\cot^2 (x)}}$

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

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

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

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

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

$=\sin (x)$
• Nov 28th 2010, 08:07 PM
Diesal
Educated, you really are educated. Thank you.
• Dec 3rd 2010, 08:57 PM
SammyS
Quote:

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?

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

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

$\displaystyle =\sqrt{1\over{1+\cot^2 x}}$
• Dec 4th 2010, 06:07 AM
Soroban
Hello, Diesal!

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

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

Quote:

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

The right side is:

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

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

and remember to reject the negative answer