Computing sin(π/5) with trigonometry and algebra

[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.