how does sin(x)cos(3pi/2)+cos(x)sin(3pi/2) become -cos(x)?
To simplify the expression sin(x)cos(3π/2) + cos(x)sin(3π/2), we can use the trigonometric identity sin(A + B) = sin(A)cos(B) + cos(A)sin(B).
In this case, A = x and B = 3π/2. We have sin(x)cos(3π/2) + cos(x)sin(3π/2) = sin(x + 3π/2).
Since sin(x + 3π/2) is equal to -cos(x) according to the unit circle, the simplified expression is indeed -cos(x).
To simplify the expression sin(x)cos(3π/2) + cos(x)sin(3π/2) and show that it equals -cos(x), we need to use the trigonometric identities.
First, let's substitute the values of cos(3π/2) and sin(3π/2):
sin(x)cos(3π/2) + cos(x)sin(3π/2)
sin(x)(0) + cos(x)(-1) (since cos(3π/2) = 0 and sin(3π/2) = -1)
0 - cos(x) = -cos(x)
Therefore, sin(x)cos(3π/2) + cos(x)sin(3π/2) simplifies to -cos(x).