Recent content by 01suite

so yea..i got the R.S..i got -1 as the answer...it looks right...but i still don't get the LS L.S = (1- sin2x)/cos2x = 1- [sin (x+x)]/[cos (x+x)] = (1- 2sinxcosx)/(2cos^2x -1) = (1- 2sinxcosx)/(1-2sin^2x) wut do i do after that ?? and one more question...wut do they mean when they...

ohh thankss I am going to go try it again thanks a lot =D

how do you put the entire denominator on 1 single fraction...like you mean... 1/1+(cosx/sinx)?

umm... L.S = (1- sin2x)/cos2x = 1- [sin (x+x)]/[cos (x+x)] = (1- 2sinxcosx)/(2cos^2x -1) = (1- 2sinxcosx)/(1-2sin^2x) ..then i don't know how to simplify that anymore then... R.S = (1-tanx)/(1+tanx) = [1-(sinx/cosx)]/[1+(sinx/cosx)] = [1-(sinx/cosx)] x [1+(cosx/sinx)] <--(?) = 1+...

i tried it but i got different answer for both sides...can someone teach me the steps on solving it. ><

ohh my badd .. the equation should be... (1-sin2x)/(cos2x) = (1-tanx)/(1+tanx) sorry//./

I have another equation that i don't know how to solve Prove: 1-sin2x/cos2x = 1-tanx/1+tanx thanks in advance to whoever that's solving it, thanks alot:smile:

umm..thanks..

the thing is...i don't even know if i did the equation right... T.T it seems wrong

soo okay...i have this question that I am stuck on... Solve for x, cos 2x - cos^2x = 0, -180degrees<x<180degrees i tried solving it... cos 2x - cos^2x = 0 [2cos^2x -1]-cos^2x = 0 cos^2x -1 = 0 <--(?) (cos x -1)(cos x +1) = 0 therefore... 1)cos x = -1 , 2)cos x = 1 so...