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after every sin and cos should be a theta, I just don't know how to type that.

2006-10-27 14:49:27 · 5 answers · asked by wc_gam3r 1 in Science & Mathematics Mathematics

5 answers

easy

let it be t
LHS = 1- (sin^2 t)/(1-cos t)
= 1-(1- cos^2 t)/(1- cos t) as sin^2t + cos^2 t = 1
= 1-(1+cos t)
= - cos t
= RHS

2006-10-27 14:54:12 · answer #1 · answered by Mein Hoon Na 7 · 0 0

Let the computer do the work......



$ mathomatic
Mathomatic version 12.5.10 (www.mathomatic.org)
Copyright (C) 1987-2006 George Gesslein II.
50 equation spaces available, 960000 bytes per equation space.

1-> 1-(sin^2 / 1-cos) = -cos

#1: 1 - (sin^2) + cos = -1*cos

1-> simplify

#1: 1 - (sin^2) + cos = -1*cos

2006-10-27 21:56:34 · answer #2 · answered by Anonymous · 0 1

x <> pi/2 + k * pi;
1+cos = sin ^ 2 / 1-cos
1 - cos ^ 2 = sin ^ 2
x <> pi/2 + k * pi is the root of this equation

2006-10-27 22:11:15 · answer #3 · answered by James Chan 4 · 0 0

Let theta = x
Then we know,
sinx^2 = 1- cosx^2 = (1-cosx)(1+cosx) -------- 1

Now using (1)
sinx^2/(1-cosx) = (1+cosx)(1-cosx)/(1-cosx) = 1+cosx

Hence LHS = 1-(1+cosx) = -cosx =RHS
and hence proved!

2006-10-27 22:29:45 · answer #4 · answered by anjali 2 · 0 0

LHS = 1 - ( sin²θ / (1 - cosθ) )
= 1 - ( (1 - cos²θ) / (1 - cosθ) )
= 1- ( (1 - cosθ) (1 + cosθ) / (1 - cosθ) )
= 1- (1+cosθ)
= 1 - 1 - cosθ
= - cosθ = RHS

2006-10-27 23:02:38 · answer #5 · answered by M. Abuhelwa 5 · 0 0

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