# Trigonometric Identities

• Apr 6th 2008, 01:35 PM
kelsey3
Trigonometric Identities
1. Verify the identity: (sinx-2+1/sinx)/(sinx-1/sinx)=sinx-1/sinx+1
2.Verify the identity: (1/1-cosx)-(cosx/1+cosx)=2csc(squared)x-1
3.Verify the identity: 2cos4xsin2x=2sin3xcos3x-2cosxsinx
• Apr 6th 2008, 01:45 PM
Mathstud28
Ok
here is what you have $\frac{sin(x)-2+\frac{1}{sin(x)}}{sin(x)-\frac{1}{sin(x)}}$..multiply top and bottom by $sin(x)$ to get $\frac{sin(x)^2-2sin(x)+1}{sin(x)^2-1}$ which simplifies to $\frac{(sin(x)-1)^2}{(sin(x)-1)(sin(x)+1)}$..and by cancelling the common factor of $sin(x)-1$ you get what you want
• Apr 6th 2008, 01:45 PM
a tutor
$\frac{sinx-2+1/sinx}{sinx-1/sinx}=\frac{sinx-1}{sinx+1}$

Working on the LHS

Multiply top and bottom by sinx and then factorise.
• Apr 6th 2008, 01:51 PM
Mathstud28
For the secodn one
$\frac{1}{1-cos(x)}-\frac{cos(x)}{1+cos(x)}$ which then by combining the fraction gives you $\frac{1+cos(x)-cos(x)+cos(x)^2}{sin(x)^2}$...the top simplifies to $\frac{2-sin(x)^2}{sin(x)^2}$ by manipulating the pythagorean identity..now seperating that gives what you wnat