Could somebody please prove this for me?
(1-cosx)/sinx=sinx/(1+cosx)
2007-02-22 05:07:31 · 4 answers · asked by arae123 1 in Science & Mathematics ➔ Mathematics
(1-cosx)/sinx = [(1-cosx)*(1+cosx)]/[sinx*(1+cosx)]
= [1- (cosx)^2]/[sinx*(1+cosx)]
=[(sinx)^2]/[sinx*(1+cosx)]
=sinx/(1+cosx)
in case you aren't sure how I got from step 2 to 3, I used the identity: (sinx)^2 + (cosx)^2 = 1, so 1 - (cosx)^2 = (sinx)^2
2007-02-22 05:15:19 · answer #1 · answered by Michael C 3 · 0⤊ 0⤋
Cross-multiply... you'll find the identity
sin²x = 1-cos²x
→ sin²x = (1-cosx)(1+cosx)
→ sinx = (1-cosx)(1+cosx)/sinx
→ sinx/(1+cosx) = (1-cosx)/sinx.
QED.
2007-02-22 05:13:24 · answer #2 · answered by Anonymous · 0⤊ 0⤋
(1-cosx)/sinx = sinx/(1+cosx) multiply across by sin(x)(1 + cos(x)) (1 - cos(x))(1 + cos(x)) = sin(x)sin(x) 1 - cos^2(x) = sin^2(x) sin^2(x) + cos^2(x) = 1 => Basic identity.
2016-05-23 23:17:45 · answer #3 · answered by Anonymous · 0⤊ 0⤋
(1-cosx)*(1+cosx)=sinx*sinx
1-cosx*cosx=sinx*sinx
clear?
2007-02-22 05:44:02 · answer #4 · answered by Anonymous · 0⤊ 0⤋