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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!!