Question

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

Answer 1

`-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)`

Answer 2

`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?