Can We Find sin(pi/60) or sin(3°)

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[@anthonycheng1765]

When theta is small, sin(theta) ~ theta. sin(pi/60) ~ pi/60

[@HATTRICK202]

It's actually in physics. In mathematics exams like entrance examination for ISIs in India, we have to find mathematical values through proper mathematics.

[@klementhajrullaj1222]

We want the exact value of sin 3° and not appromaxtly value!

[@Ayush-yj5qv]

Fk approximations

[@ishansh0077]

unsatisfying result.

[@sagarmajumder7806]

(Theta is small) implies (theta->0). Now you tell is ( 3->0) true? As it is false, so your approximation doesn't work here.

[@XJWill1]

Here is another way to express the number: sind( 3) = 1/4 *sqrt( 8-sqrt(15)-sqrt(10-2*sqrt(5))-sqrt(3) )

[@alnitaka]

This reminds me of a paper in the College Mathematics Journal, 2016 November, entitled "The Sine of a Single Degree". The first part looks like your video, and then it says you can't get an expression involving real radicals for the sine of one degree, because it involves solving a cubic equation.

[@SyberMath]

I’d like to see that article but I guess I need jstor access 😔

[@Khashayarissi-ob4yj]

With luck and more power to you.

[@SyberMath]

❤️

[@Nobodyman181]

Thanks!

[@SyberMath]

No problem!

[@Qermaq]

While we're on trig, maybe this would be fun to show: if a/b is sufficiently close to being appx. sqrt(n), then (a + bn)/(a + b) is even closer,, and iterating this function tends toward sqrt(n).

[@SyberMath]

Interesting!

[@finnboltz]

How can we prove that?

[@Qermaq]

@@finnboltz That's just the thing - I don't know! I'm assuming that since this resembles a trig function maybe there's a trig way to prove it.

[@scottleung9587]

Nice!

[@SyberMath]

Thanks!

[@Qermaq]

Here's an eccentric method. The PRT 57, 176, 185 has an acute angle of about 18 degrees. The PRT 69,260, 269 has an acute angle of about 15 degrees. So an approximation of sine of 3 degrees can be derived as (57/185)(260/269) - (176/185)(69/269) which yields another PRT 2676, 49693, 49765 which has acute angle of about 3 degrees.

[@BukhalovAV]

What is PRT?

[@Qermaq]

@@BukhalovAV Pythagorean Right Triangle.

[@phill3986]

😊😊😊👍👍👍

[@Chrisoikmath_]

Syber vs Geometry 1-0 🤩

[@edding8400]

Make triangles great again

[@SyberMath]

Yess! MTGA

[@gregwochlik9233]

To approximate the value, use an aviatiors trick. For small angles (emphasis on small), sin (1) = 1/60. Multiply by 3 to get sin(3) to be around 1/20. or 0.05.

[@MortezaSabzian-db1sl]

π/60*(1-1/6*(π/60)²)

[@klementhajrullaj1222]

How to find sin 23°? Thank you!

[@HATTRICK202]

Here, we saw sin3°. Use sin60° and take 60°=3x and find sinx means sin20°. Then use sin(20°+3°) = sin20°cos3°+ cos20°sin3°.

[@XJWill1]

If the degree value is not a multiple of 3, then the sine of the degree value will not be expressible in terms of real-valued arithmetic and square roots. It requires cube roots and arithmetic with imaginary numbers. sind(23)= 1/sqrt(1-(-1+16/(8-(sqrt(48)+sqrt(-16))^(1/3)*(1+sqrt(5)+sqrt(-10+sqrt(20)))))^2)

[@XJWill1]

In case it was not apparent from my previous comment, an expression involving cube roots and complex-valued numbers is not useful for computing trig functions of non-multiple-of-3-degrees angles. The reason is that computing the complex-valued cube root is equivalent to numerically evaluating the sine and cosine (or tangent) of the arguments since extracting the cube root involves converting the complex-valued number to polar form.

[@XJWill1]

The square root of a complex-valued number can be expressed in terms of square roots and real-valued arithmetic. But the cube root of a complex-valued number cannot be expressed without involving trig functions or complex exponentials and logarithms.

[@SyberMath]

How did you find it? You gotta teach me 🤓

[@DonEnsley-mathdrum]

problem sin(π/60) = ? I took sin(75°-72°) = (¹/₁₆){ (√5-1)[√3 √(2+√3)+ √(2-√3)] - √(10+2√5)[√(2+√3)-√3 √(2-√3)] } 75° = 60°+15° = 60°+ ½ 30° sin 72° = ½ √(10+2√5) cos 72° = ½ √(√5-1) Using the sum and difference formulas, and the half angle formulas, the result is (answer) sin(π/60)= {(√5-1)[√3 √(2+√3)+√(2-√3)] - √(10+2√5)[√(2+√3)-√3 √(2-√3)]}/16 (approx. 0.052335956242904)

[@SyberMath]

That’s so cool!

[@DonEnsley-mathdrum]

@@SyberMath apparently 75-72 still equals 3! Lol

[@BukhalovAV]

Why don't you use degree mark when you mean degree? It means radians out of context, and this is a rude failure.

[@SyberMath]

I don’t like writing it! It should be understood 🤪