Simplify this trig expression to +- cosx or +-sinx?

sin(3π/2)cos(x) + cos(3π/2)sin(x) =

-1(cos(x)) + 0sin(x) = -cos(x)

3/2pi is not really the way to show what you are trying to show.

I would go with :-

sin [ 3π/2 + x ]

sin (3π/2) cos x + cos (3π/2) sin x

cos x + 0

cos x

{if you are familiar with this formula then follow thelink suggested by ananymousPerson}

sin(A+B)=sin(A)cos(B)+sin(B)cos(A)

Putting A=3á´¨/2 and B=x,

sin(3á´¨/2 + x)=sin(3á´¨/2)cos(x)+sin(x)cos(3á´¨/2)

sin(3á´¨/2)=sin(-á´¨/2)=-1

cos(3á´¨/2)=cos(-á´¨/2)=0, so

sin(3á´¨/2 + x)=-cos(x)

first, remember your standard trig identities

sin(a + b) = sin(a)cos(b) + sin(b)cos(a)

so

your starting point

==> sin(3π/2)cos(x) + sin(x)cos(3π/2)

now, sin(3π/2) = - sin(π/2) = -1

and cos(3π/2) = cos(π/2) = 0

so

==> -cos(x) + 0

==> -cos(x)

and you are done

Check out the angle addition formulas on the first page of the site below.

sin(a + b) = sinacosb + cosa sinb

sin(3/2pi + x) = sin3/2p*cosx +cos3/2pi*sinx

= - sinx

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