# How do you find the exact value of sin24 using the sum and difference, double angle or half angle formulas?

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Dean R. Share
Jun 8, 2018

$\sin {24}^{\circ} = \frac{1}{8} \left(\sqrt{3} + \sqrt{15} - \sqrt{10 - 2 \sqrt{5}}\right)$

#### Explanation:

$\sin {24}^{\circ}$

Angles which are multiples of three degrees are constructible with a straightedge and compass, and thus have a closed form consisting of integers combined by addition, subtraction, multiplication, division and square roots (of positive numbers).

This one is doable if we know the trig functions of ${144}^{\circ}$ because ${24}^{\circ} = {144}^{\circ} - {120}^{\circ}$ and ${120}^{\circ}$ is of course related to one of the Two Tired Triangles of Trig, 30/60/90.

I happen to know $\cos {144}^{\circ} = - \frac{\phi}{2}$, negative one-half the Golden Ratio. Let's derive that.

In general $\cos 3 \theta = \cos 2 \theta$ has solutions

$3 \theta = \pm 2 \theta + {360}^{\circ} k \quad$ integer $k$

The minus sign subsumes the plus, and we end up with

$5 \theta = {360}^{\circ} k$

$\theta = {72}^{\circ} k$

If we let $x = \cos \theta$ then using the double and triple angle formulas $\cos 3 \theta = \cos 2 \theta$ becomes

$4 {x}^{3} - 3 x = 2 {x}^{2} - 1$

$4 {x}^{3} - 2 {x}^{2} - 3 x + 1 = 0$

Since $\theta = {72}^{\circ} k$ we know $\theta = 0$ works, so $x = \cos 0 = 1$ is a root of this polynomial. We factor:

$\left(x - 1\right) \left(4 {x}^{2} + 2 x - 1\right) = 0$

The quadratic has solutions

$x = \frac{1}{4} \left(- 1 \pm \sqrt{5}\right)$

$\cos {144}^{\circ}$ is the only negative root of the original cubic. $\cos {0}^{\circ}$ and $\cos {72}^{\circ}$ are the other two roots, both positive.

$\cos {144}^{\circ} = - \frac{1}{4} \left(1 + \sqrt{5}\right)$

Like I said, negative one half the Golden Ratio $\phi = \frac{1 + \sqrt{5}}{2.}$

$\sin {144}^{\circ}$ is positive. Unfortunately it has a nested square root; we live with it.

$\sin {144}^{\circ} = \sqrt{1 - {\cos}^{2} {144}^{\circ}} = \sqrt{1 - \frac{1}{16} \left(6 + 2 \sqrt{5}\right)} = \setminus \sqrt{\frac{1}{8} \left(5 - \sqrt{5}\right)}$

Now we can finally apply the difference angle formula:

$\sin {24}^{\circ} = \sin \left({144}^{\circ} - {120}^{\circ}\right)$

$= \sin {144}^{\circ} \cos {120}^{\circ} - \cos {144}^{\circ} \sin {120}^{\circ}$

$= \sqrt{\frac{1}{8} \left(5 - \sqrt{5}\right)} \left(- \frac{1}{2}\right) - \left(- \frac{1}{4} \left(1 + \sqrt{5}\right)\right) \left(\frac{\sqrt{3}}{2}\right)$

= -1/8 sqrt{16/8(5-sqrt{5}) } + 1/8(sqrt{3} + sqrt{15}))

$\sin {24}^{\circ} = \frac{1}{8} \left(\sqrt{3} + \sqrt{15} - \sqrt{10 - 2 \sqrt{5}}\right)$

Check: Calculator.

 sin 24^circ - 1/8 (sqrt(3) + sqrt(15) - sqrt(10 - 2 sqrt(5))) =0 quad sqrt

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