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

Question

Madelyn Adams · Accepted Answer

#-cos(pi/5)# 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)#

Brooklyn Anderson · Answer

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

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

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?