prove this identity: (cos x/2 - sin x/2)^2 =1-sin x?

Answer 1

Expand the exponent: (cos (x/2) - sin(x/2))^2 = (cos x/2)^2 - 2*cos(x/2)*sin(x/2) + (sin x/2)^2 = 1 - 2*sin(x/2)*cos(x/2) = 1-sin(x)

using the fact (cos a)^2 + (sin a)^2 = 1 and 2*sin(a)*cos(a) = sin(2a)

Steve

Answer 2

Remember
cos x/2, aka cos 1/2x, is sqrt(1/2 + 1/2cos x)
sin x/2, aka sin 1/2x, is sqrt(1/2-1/2cos x)

Rewrite the left side of the identity
[sqrt(1/2+1/2cos x) - sqrt(1/2-1/2cos x)]^2
Simplify
[sqrt(1/2+1/2cos x) - sqrt(1/2-1/2cos x)][sqrt(1/2+1/2cos x) - sqrt(1/2-1/2cos x)]

(1/2+1/2cos x) -(sqrt[1/2+1/2cos x)(1/2-1/2cos x)] - (sqrt[1/2+1/2cos x)(1/2-1/2cos x)] + (1/2-1/2cosx)

1 - 2(sqrt[1/2+1/2cos x)(1/2-1/2cos x)]
1 - 2(sqrt[1/4 -1/4cos^2x + 1/4cos^2x -1/4cos^2x]
1 - 2(sqrt[1/4 - 1/4cos^2x]
1 - 2(sqrt[1/4(1 - cos^2x]
1 - 2(1/2sqrt[1-cos^2x]
1- sqrt[1 - cos^2x]
1 - sqrt[sin^2x]
1-sin x

Answer 3

www.wikipedia.org/wiki/Trig_identities

try the half angle and pythagorean identities

i tried to solve it myself but its been a while since ive done proofs so i got a totally crazy answer. good luck!

Answer 4

No.