how do you prove... 1-(sin^2 / 1-cos) = -cos?

Question

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

Answer 1

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

Answer 2

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



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

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

1-> simplify

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

Answer 3

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

Answer 4

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!

Answer 5

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