How do you solve #sin(pi/5-pi/2)#?

Expand use Difference of Angles.
#sin(pi/5 - pi/2) = sin(pi/5)cos(pi/2) - cos(pi/5)sin(pi/2)#

Remember #cos(pi/2) = 0 and sin(pi/2) = 1#
So we can simplify it more:
#= sin(pi/5)*0 - cos(pi/5)*1= - cos(pi/5)#

Finally a triangle that's not 30/60/90 or 45/45/90. Thank you.

This is our first tiptoe outside the two tired triangles of trig. We can find a nice radical expression for this one, which is related to the pentagon. It's a bit involved, but not too bad.

Let's do it in degrees, which always makes more sense to me.

# pi/5 = 36^circ quad quad pi/2=90^circ #

#sin(pi/5 - pi/2) = sin(36^circ - 90^circ) = cos(90 - (36^circ - 90^circ)) = cos(180^circ-36^circ) = cos 144^circ #

Strap in, this is where it gets interesting. First we note #cos(3 theta)=cos(2 theta)# has solutions #3 theta = pm 2 theta + 360^circ k,# integer #k#, or just using the subsuming minus sign,

#theta = 72 ^circ k#

In other words, all the multiples of #72^circ,# including #144^circ# are solutions to #cos3 theta = cos 2 theta.# Let's write the triple and double angle formulas,

#4 cos ^3 theta - 3 cos theta = 2 cos ^2 theta - 1 #

Our unknown now is #x=cos theta,# because #cos 144^circ# is among the roots.

#4 x^3 - 3 x = 2 x^2 - 1 #

# 4x^3 - 2 x^2 -3x + 1 = 0 #

It's a cubic, but an easy one because we know #x = cos (0 times 72^circ)=1# must be a root, so #x-1# must be a factor:

# (x-1) ( 4 x^2 + 2 x - 1) = 0#

The quadratic has roots

# x = 1/4 (-1 pm sqrt{5})#

Clearly #cos 72^circ# is the positive root and #cos 144^circ# is the negative root.

# cos 144^circ = 1/4 (-1 -sqrt{5})#

#sin(pi/5 - pi/2) = cos 144^circ = 1/4 (-1 - sqrt{5})#

That's #-1/2# of the Golden Ratio.

Now that wasn't too bad venturing beyond our two tired triangles of trig, was it?