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

2 Answers
May 11, 2018

-cos(pi/5)

Explanation:

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)

May 11, 2018

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

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

Explanation:

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?