ImDaStyle Full Member 10+ Year Member 15+ Year Member Joined May 15, 2008 Messages 65 Reaction score 0 May 15, 2008 #1 This one problem is driving me insane, please help... =D Verify! cot^3x -------- = cosx (csc^2x-1) cscx Members don't see this ad.
This one problem is driving me insane, please help... =D Verify! cot^3x -------- = cosx (csc^2x-1) cscx Members don't see this ad.
bigstix808 Mac Daddy Member 15+ Year Member Joined Oct 10, 2007 Messages 1,079 Reaction score 34 May 15, 2008 #2 Upvote 0 Downvote
S Streetwolf Ultra Senior Member Verified Member 10+ Year Member Dentist 15+ Year Member Joined Oct 25, 2006 Messages 1,801 Reaction score 7 May 15, 2008 #3 ImDaStyle said: This one problem is driving me insane, please help... =D Verify! cot^3x -------- = cosx (csc^2x-1) cscx Click to expand... cot(x) = cos(x)/sin(x) csc(x) = 1/sin(x) So [cos^3(x) / sin^3(x)] / (1/sin(x)) = cos^3(x) / sin^2(x) cos^3(x) = cos(x)cos^2(x) = cos(x)*[1-sin^2(x)] = cos(x) - cos(x)sin^2(x) Divide all that by sin^2(x) and you get cos(x)/sin^2(x) - cos(x). That is the same as cos(x) * [1/sin^2(x) - 1]. That is the same as cos(x) * [csc^2(x) - 1]. Upvote 0 Downvote
ImDaStyle said: This one problem is driving me insane, please help... =D Verify! cot^3x -------- = cosx (csc^2x-1) cscx Click to expand... cot(x) = cos(x)/sin(x) csc(x) = 1/sin(x) So [cos^3(x) / sin^3(x)] / (1/sin(x)) = cos^3(x) / sin^2(x) cos^3(x) = cos(x)cos^2(x) = cos(x)*[1-sin^2(x)] = cos(x) - cos(x)sin^2(x) Divide all that by sin^2(x) and you get cos(x)/sin^2(x) - cos(x). That is the same as cos(x) * [1/sin^2(x) - 1]. That is the same as cos(x) * [csc^2(x) - 1].
ImDaStyle Full Member 10+ Year Member 15+ Year Member Joined May 15, 2008 Messages 65 Reaction score 0 May 15, 2008 #4 Yay ^__^ Thanks so much! I understand the problem at last xD! Upvote 0 Downvote