S: Math Help - Solving Trig Quiz Problem

[dextercioby]

I proposed this problem on a thread in the homework section.I honestly do not know the answer...

"Compute \sin\frac{\pi}{5} using trigonometry and algebra only."


Daniel.

 

[Curious3141]

It's actually fairly simple. I haven't worked it all the way out to the bitter end, but I've verified my method and it works.

Expand \sin 5\theta = \sin(3\theta + 2\theta) in terms of \sin \theta. You would just use the addition formula followed by expanding each of the terms with the triple and double angle formulae.

All the terms with powers of \cos\theta will have even powers of the cosine, so they can easily be converted to even powers of sine with \cos^2\theta = 1 - \sin^2\theta.

You will eventually get a reducible quintic in terms of sine theta. Let s = \sin\theta

\sin 5\theta = 16s^5 - 20s^3 + 5s

Equate that to zero (since \sin\pi = 0) and solve.

Dismissing s = 0, it becomes a quartic which is actually a quadratic in s^2, which you can solve to get :

s^2 = \frac{1}{8}(5 \pm \sqrt{5})

One of the values (with the plus sign is a redundant root).

EDIT : I've not yet found a way to determine the explicit value of s from that expression. I keep getting ugly expressions with more roots of surds. I'm still working on this part, but for now, my answer is :

\sin{\frac{\pi}{5}} = \sqrt{\frac{1}{8}(5 - \sqrt{5})}

 

[Curious3141]

And, as a matter of fact,

\cos{\frac{\pi}{5}} = \frac{1}{2}\phi

where \phi = \frac{1}{2}(1 + \sqrt{5}), the golden ratio. But the expression for the sine still involves squaring that, subtracting the result from unity, and taking the root, and I cannot find a way to simplify that.