# Thread: Difficult Trig Identity Question

1. ## 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?

2. 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 $\displaystyle \sin (x) = \sqrt{1+\cot^2 (x)}$?? 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.

3. How do I write it the way you did it?

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

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

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

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

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

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

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

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

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

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

$\displaystyle =\sin (x)$

8. Educated, you really are educated. Thank you.

9. 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?

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

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

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

10. Hello, Diesal!

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

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

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

The right side is:

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

11. You should end off with $\displaystyle \sqrt{sin^x}=sinx$ OR $\displaystyle -sinx$

and remember to reject the negative answer