Find all x, 0 < x < pi, such that sin^2x=1/2

sin x = 1/ sqrt(2) or - 1/sqrt(2)

x= pi/4, 3pi/4 , 5pi/4 or 7pi/4 (1st 2 for 1/sqrt(2) and next 2 for - 1/sqrt(2)

If A, B, C are the angles of a triangle...?

prove that cos [(B-C) / 2] - sin (A/2) = 2 sin (B/2) sin (C/2)

A+B+C = pi

so A/2 = pi-(B+C)/2

sin (A/2) = sin (pi-(B+C)/2) = cos (B+C)/2

so
cos [(B-C) / 2] - sin (A/2) =

= cos [(B-C) / 2] - cos [(B+C)/2] = 2 sin (B/2) sin (C/2)

using COS A - COS B = 2 cos sin (A+B)/2 sin (A-B)/2

simplify the following?

simplify the following? cos(2x-y)cosy - sin(2x-y)siny?

using cos A cos B - sin A sin B = cos(A+B) and putting A = 2x-y and B = y we get

cos(2x-y)cosy - sin(2x-y)siny = cos (2x-y+y) = cos 2x