Expert: Sherry Wallin - 11/5/2010

Question
I have been frying my brain trying to figure this equation. verify that:

(cos(x)/2 - sin(x)/2)^2 = 1-sinx

Answer
Adam~

Maybe the statement is not true if you can't "figure it out". You always want to believe something can be proven before you waste your time trying to proof something that canNOT be proven.

Let x = 0
[(cos 0)/2 - (sin0)/2]^2 = [1/2 - 0] = [1/2]^2 = 1/4
and 1 - sin 0 = 1 - 0 = 1 and 1/2 ~= 1


Math Prof


PS  If you meant [cos(x/2) - sin(x/2])^2 = 1 - sinx then you can prove it as follows

let y = x/2 (this substitution makes it easier to prove)

[cos(x/2) - sin (x/2)]^2 = (cosy - siny)^2 and 1 - sin x is 1 - sin 2y
(cos(y) - sin(y))^2 = [cos(y)]^2 - 2sin(y)cos(y) + [sin(y)]^2
= [cos(y)]^2 + [sin(y)]^2 - 2sin(y)cos(y)
= 1 - 2sin(y)cos(y)
= 1 - sin(2y)

and now substitute back y = x/2 so 1 - sin(2y) = 1 - sin(2(x/2))
= 1 - sin x   as desired...

Notice if what you meant is the 2nd way I did it you needed to use the parentheses as I did because what you provided did NOT mean the same thing...