Finding the EXACT Values for sin(10), sin(50), and sin(70) Using the Cubic Formula

This video works to determine the exact values for the sine of 10 degrees, sin(10), the sine of 50 degrees, sin(50), and the sine of 70 degrees, sin(70). It does so by writing a trigonometric equation, changing it to a cubic polynomial equation, investigating its roots, and finally using Cardano's Cubic Formula to determine the three values. It also shows that despite utilizing complex numbers, the exact values are irrational in nature.

Writing the Equation: 1:26
Trying the Rational Root Theorem: 3:03
Looking at the Graph of the Function: 3:31
Using The Cubic Formula: 6:18
The Three Exact Values: 11:20
Finding the Product of the Three Values: 11:31

Cubic Formula:

[johnspence]

Nice clear step by step, its nice to see a video that doesnt say "as proven in an earlier video" and just skip massively important steps. Great stuff, which all math profs were this clear.

[Простонякойпъпеш]

I dont know how to say it, but so far your the best YouTube math teacher iv seen.U gave me EXACTLY what is war looking for😀😀😀.Amazing

[SUBSUN]

“Our narrative seems to have turned into a dark turn indeed” you just summed up my life right now.

[benkimball]

Simply beautiful, the way the world of mathematics works will always remain an enigma to me, how the ugliness can work together and create something so, as you said, exquisite. Well done good sir, you have earned yourself a subscription for your work.

[XJWill]

Here are my preferred forms for the exact values of sine of 10, 20, 40, 50, 70, and 80 degrees. I have written the equations so that they can be copied and pasted into Wolfram Alpha for verification. Also, the right-hand sides of the equations are written so they can be parsed by Wolfram Alpha, Octave, or Maxima.

sin(10deg) = sqrt( 16^(1/3) - (-sqrt(-3)+1)^(1/3) - (sqrt(-3)+1)^(1/3) )/128^(1/6)
sin(20deg) = ( (-sqrt(-3)+1)^(1/3) - (sqrt(-3)+1)^(1/3) ) * sqrt(-1)/(16)^(1/3)
sin(40deg) = ( (-sqrt(-3)-1)^(1/3) - (sqrt(-3)-1)^(1/3) ) * sqrt(-1)/(16)^(1/3)
sin(50deg) = ( (-sqrt(-3) - 1)^(1/3) + (sqrt(-3) - 1)^(1/3) ) / 16^(1/3)
sin(70deg) = ( (-sqrt(-3) + 1)^(1/3) + (sqrt(-3) + 1)^(1/3) ) / 16^(1/3)
sin(80deg) = sqrt( 16^(1/3) + (-sqrt(-3)+1)^(1/3) + (sqrt(-3)+1)^(1/3) )/128^(1/6)

I'm looking forward to your next video for computing the values of sin(20deg), sin(40deg), and sin(80deg).

[XJWill]

Here are some even shorter forms for the multiples of 5° that are not multiples of 3°. If I understand the rules of the game, they are to come up with expressions for the sine of angles that use only arithmetic operations, square roots, and cube roots. If higher roots than cube roots were allowed, then it is almost trivial to write a simple expression using the 180th root together with Euler's identity. But since we want to increase the difficulty, we limit it to square roots and cube roots. So here are the shortest forms I know of that satisfy those criteria:

sind( 5)= 1/sqrt(1-(1-4/(2-(
sind(10)= 1/sqrt(1-(1-4/(2-( 4+sqrt(-48))^(1/3)))^2)
sind(20)=
sind(25)=
sind(35)= 1/sqrt(1-(1-4/(2-(
sind(40)=
sind(50)=
sind(55)= 1/sqrt(1-(1-4/(2+(
sind(65)=
sind(70)=
sind(80)= 1/sqrt(1-(1-4/(2+( 4+sqrt(-48))^(1/3)))^2)
sind(85)= 1/sqrt(1-(1-4/(2+(

You can copy and paste those into Wolfram Alpha to get a better visual representation. They can also be parsed by Octave or Maxima.

Most of those were derived by first finding the tangent of the corresponding angle (or complement of the angle). This can be done by writing tan(x) in terms of complex exponentials and manipulating the equation so there is only one complex exponential of the form exp(2*i*x), then raise that to the 3rd power and then take the cube root so that we essentially have a sixth-angle formula for tan(). So tand(5) is calculated with 6*5=30 degrees of sin() and cos(), with only a single cube root in the expression. Then the trig identity sec^2 - tan^2 = 1 is used to find cos = 1/sec, and then sind(x) comes from cosd(90-x).

[Nirajthakur]

Sir Really You are A Good techer ..🙏🙏
Sir Love from India...🇮🇳🇮🇳🇮🇳🇮🇳

[kennybutcher]

holy crap... What? Did I just learn... Thanks James!

[zanti]

There is a neat shortcut to find sin 10⁰ * sin 50⁰ * sin 70⁰: From your previous work, you know the roots of 8x³ -6x + 1 = 0 are sin 10⁰, sin 50⁰, and sin -70⁰. By Vieta's Formula, the product of these roots is -1/8. Now just note that sin 70⁰ = -sin -70⁰, therefore sin 10⁰ * sin 50⁰ * sin 70⁰ = - (-1/8) = 1/8, and you're done.

[kjellsandler]

How did Mathematicians come up with the formulas for sine and cosine of x/3?

[jarikosonen]

This looks good, but is it impossible to get the imaginary numbers removed from the result of sin(10), sin(50) & sin(70)?
It might be quite hard since they are under the radical cubic both...
This looks to be possible almost for any angle to get the exact value for that angle.

[Amoeby]

Can we somehow express sin of 10 degrees in radicals?

[HUNTER-qnff]

Again pls sir, at last why -1/8 become to 1/8 ( i don't get, my fault sir)

[zanti]

Now this is interesting... I tried simplifying the cubic by taking out a -1/2 instead of a +1/2 and ended up with (³√((1+i√3)/2) + ³√((1-i√3)/2))/2, which comes out to sin 70⁰. This exact value for sin 70⁰ looks almost identical to the exact value for sin 50⁰ in the video, the only different being a pair of signs in the cube root are now + instead of -. I'd imagine the value for sin 70⁰ shown in the video can be simplified to this value somehow.

[dugong]

You might as well say and there you have your value in terms of complex numbers. Can it be expressed in terms of elementary functions of real numbers? Unfortunately my hopes of learning this information by watching your video were dashed to smithereens. On the plus side, you have a pleasant voice and manner of speaking!