Evaluate sin 20 ?

1 Answer
Øko
Feb 22, 2018

#sin(20^@)~~0.34202014332566# 14 decimal places

Explanation:

First method:

And by far the easiest method is to use a calculator

#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(3theta)=3sin(theta)-4sin^3(theta)#

Let #theta=20^@#

#sin(60^@)=3sin(20^@)-4sin^3(20^@)#

But #sin(60^@)=sqrt(3)/2#

#sqrt(3)/2=3sin(20^@)-4sin^3(20^@)#

#=>3sin(20^@)-4sin^3(20^@)-sqrt(3)/2=0#

Let #x=sin(20^@)#

#3x-4x^3-sqrt(3)/2=0#

#3/4x-x^3-sqrt(3)/8=0#

#x^3-3/4x+sqrt(3)/8=0#

In other words #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

Related questions
Impact of this question
7547 views around the world
You can reuse this answer
Creative Commons License