how is sin(x-3π/2)=cosx?
could you please explain?
could you please explain?
-
for any angles, a & b, you have:
sin(a - b) = sin(a)cos(b) - cos(a)sin(b)
substitute x for z and 3π/2 for b
sin(x-3π/2) = sin(x)cos(3π/2) - cos(x)sin(3π/2)
But cos(3π/2) = 0 and sin(3π/2) = -1 Therefore:
sin(x - 3π/2) = sin(x)*0 - cos(x)*(-1) = 0 + cos(x) = cos(x)
sin(a - b) = sin(a)cos(b) - cos(a)sin(b)
substitute x for z and 3π/2 for b
sin(x-3π/2) = sin(x)cos(3π/2) - cos(x)sin(3π/2)
But cos(3π/2) = 0 and sin(3π/2) = -1 Therefore:
sin(x - 3π/2) = sin(x)*0 - cos(x)*(-1) = 0 + cos(x) = cos(x)