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?
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?
Learn LaTeX. You can download the Latex.pdf tutorial and look at the basic commands.
$\displaystyle \sqrt{\dfrac{1}{1+\cot^2 (x)}}$
$\displaystyle =\sqrt{\dfrac{1}{1+\frac{\cos^2 (x)}{\sin^2 (x)}}}$
$\displaystyle =\sqrt{\dfrac{1}{\frac{\sin^2 (x)}{\sin^2 (x)}+\frac{\cos^2 (x)}{\sin^2 (x)}}}$
$\displaystyle =\sqrt{\dfrac{1}{\frac{\cos^2 (x) + sin^2 (x)}{\sin^2 (x)}}}$
$\displaystyle =\sqrt{\dfrac{1}{\frac{1}{\sin^2 (x)}}}$
$\displaystyle =\sqrt{\sin^2 (x)}$
$\displaystyle =\sin (x)$
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 $