# cotangent identity problem

• Dec 26th 2009, 03:32 PM
Savior_Self
cotangent identity problem
Honestly, I'm not even sure how to start..

$Show: \frac{2cotx}{cot^2 + 1} = sin2x$

• Dec 26th 2009, 04:02 PM
galactus
Note the identity $cot^{2}(x)+1=csc^{2}(x)$

$\frac{2cot(x)}{cot^{2}(x)+1}=\frac{2\frac{cos(x)}{ sin(x)}}{csc^{2}(x)}=\frac{2\frac{cos(x)}{sin(x)}} {\frac{1}{sin^{2}(x)}}$

$=\frac{2cos(x)}{sin(x)}\cdot \frac{sin^{2}(x)}{1}=2cos(x)sin(x)=sin(2x)$
• Dec 26th 2009, 04:14 PM
Raoh
Quote:

Originally Posted by Savior_Self
Honestly, I'm not even sure how to start..

$Show: \frac{2cotx}{cot^2 + 1} = sin2x$

$\frac{2cot(x)}{cot^2(x) + 1}= \frac{2\frac{1}{\tan(x)}}{\frac{1}{\tan^2(x)}+1}$= $\frac{2\frac{\cos(x)}{\sin(x)}}{\frac{\cos^2(x)}{\ sin^2(x)}+1}$= $\frac{2\frac{\cos(x)}{\sin(x)}}{\frac{\cos^2(x)+\s in^2(x)}{\sin^2(x)}}$= $\frac{2\frac{\cos(x)}{\sin(x)}}{\frac{1}{\sin^2(x) }}$= $2\frac{\cos(x)}{\sin(x)}\times\sin^2(x)=2\cos(x)\s in(x)=\sin(2x).$
• Dec 26th 2009, 04:17 PM
Raoh
Quote:

Originally Posted by galactus
Note the identity $cot^{2}(x)+1=csc^{2}(x)$

$\frac{2cot(x)}{cot^{2}(x)+1}=\frac{2\frac{cos(x)}{ sin(x)}}{csc^{2}(x)}=\frac{2\frac{cos(x)}{sin(x)}} {\frac{1}{sin^{2}(x)}}$

$=\frac{2cos(x)}{sin(x)}\cdot \frac{sin^{2}(x)}{1}=2cos(x)sin(x)=sin(2x)$

err(Worried) i think i chose the hard way.