View Full Version : Trig identity
skr_girl
Mar 26, 2007, 01:33 PM
HELP - I am really bad at trig and need some assistance... I have to verify
sin X
______ = csc X - cot anx
1+ cos X
Can anyone help me out - give me a place to start??
Thanks
SR
galactus
Mar 26, 2007, 03:54 PM
Multiply top and bottom by 1-cos(x):
\frac{sin(x)}{1+cos(x)}\cdot\frac{1-cos(x)}{1-cos(x)}
\frac{sin(x)(1-cos(x))}{\underbrace{1-cos^{2}(x)}_{\text{sin^2(x)}}}
\frac{sin(x)(1-cos(x))}{sin^{2}(x)}
\frac{1-cos(x)}{sin(x)}=\underbrace{\frac{1}{sin(x)}}_{\te xt{csc(x)}}-\underbrace{\frac{cos(x)}{sin(x)}}_{\text{cot(x)}}
darkenedsoul24
Mar 26, 2007, 05:51 PM
first multiply both sides by 1 + cosx
then change csc x to 1/sinx and cotx to cosx/sinx
then multiply both sides by sinx
then use the fact that (sinx)^2 + (cosx)^2 = 1
skr_girl
Mar 27, 2007, 04:42 AM
Thank you ver much for your help!!