Evaluate sin 20 ?

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Evaluate sin 20 ?

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Answer 1 · 2018-02-22T11:26:46

$sin (20^{\circ}) \approx 0.34202014332566$ $sin(20^@)~~0.34202014332566$ 14 decimal places

Explanation:

First method:

And by far the easiest method is to use a calculator

$sin (20^{\circ}) \approx 0.34202014332566$ $sin(20^@)~~0.34202014332566$ 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

$sin (3 θ) = 3 sin (θ) - 4 {sin}^{3} (θ)$ $sin(3theta)=3sin(theta)-4sin^3(theta)$

Let $θ = 20^{\circ}$ $theta=20^@$

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

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

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

$\Rightarrow 3 sin (20^{\circ}) - 4 {sin}^{3} (20^{\circ}) - \frac{\sqrt{3}}{2} = 0$ $=>3sin(20^@)-4sin^3(20^@)-sqrt(3)/2=0$

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

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

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

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

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

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

$x_(n+1)=x_n-f(x_n)/(f'(x_n))$

We know

$f(x)=x^3-3/4x+sqrt(3)/2$ and $f'(x)=3x^2-3/4$

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

$x_0=1/3$

By Newton's method

$x_1=1/3-((1/3)^3-3/4(1/3)+sqrt(3)/2)/(3(1/3)^2-3/4)~~ 0.341837464493$

$x_2=x_1-f(x_1)/(f'(x_1))~~ 0.342020057633$

$x_3=x_2-f(x_2)/(f'(x_2))~~ 0.342020143326$

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

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Evaluate sin 20 ?

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