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# Evaluate sin 20 ?

Feb 22, 2018

$\sin \left({20}^{\circ}\right) \approx 0.34202014332566$ 14 decimal places

#### Explanation:

First method:

And by far the easiest method is to use a calculator

$\sin \left({20}^{\circ}\right) \approx 0.34202014332566$ 14 decimal places

Second method:

If all the trigonometric buttons of your calculator are broken,
after all the wild math :), there is another solution

Use the identity

• $\sin \left(3 \theta\right) = 3 \sin \left(\theta\right) - 4 {\sin}^{3} \left(\theta\right)$

Let $\theta = {20}^{\circ}$

$\sin \left({60}^{\circ}\right) = 3 \sin \left({20}^{\circ}\right) - 4 {\sin}^{3} \left({20}^{\circ}\right)$

But $\sin \left({60}^{\circ}\right) = \frac{\sqrt{3}}{2}$

$\frac{\sqrt{3}}{2} = 3 \sin \left({20}^{\circ}\right) - 4 {\sin}^{3} \left({20}^{\circ}\right)$

$\implies 3 \sin \left({20}^{\circ}\right) - 4 {\sin}^{3} \left({20}^{\circ}\right) - \frac{\sqrt{3}}{2} = 0$

Let $x = \sin \left({20}^{\circ}\right)$

$3 x - 4 {x}^{3} - \frac{\sqrt{3}}{2} = 0$

$\frac{3}{4} x - {x}^{3} - \frac{\sqrt{3}}{8} = 0$

${x}^{3} - \frac{3}{4} x + \frac{\sqrt{3}}{8} = 0$

In other words $\sin \left(20\right)$ must be a solution to this cubic

By Newtons's method we can approximate this root
(However be a little careful)

${x}_{n + 1} = {x}_{n} - f \frac{{x}_{n}}{f ' \left({x}_{n}\right)}$

We know

$f \left(x\right) = {x}^{3} - \frac{3}{4} x + \frac{\sqrt{3}}{2}$ and $f ' \left(x\right) = 3 {x}^{2} - \frac{3}{4}$

Drawn $\sin \left(20\right)$ looks about a third (actually a really good guess)

${x}_{0} = \frac{1}{3}$

By Newton's method

${x}_{1} = \frac{1}{3} - \frac{{\left(\frac{1}{3}\right)}^{3} - \frac{3}{4} \left(\frac{1}{3}\right) + \frac{\sqrt{3}}{2}}{3 {\left(\frac{1}{3}\right)}^{2} - \frac{3}{4}} \approx 0.341837464493$

${x}_{2} = {x}_{1} - f \frac{{x}_{1}}{f ' \left({x}_{1}\right)} \approx 0.342020057633$

${x}_{3} = {x}_{2} - f \frac{{x}_{2}}{f ' \left({x}_{2}\right)} \approx 0.342020143326$

After 3 steps precise with at least 12 decimal places

After just 6 steps, we should have a precision around 100 digits,
according to Wolfram Alpha, by Newton's Method