Calculating sin(π/5) Using a Regular Pentagon

The exact value of sin(𝜋/5) is (sqrt(10–2sqrt(5))/4. 😎
We can find it using a regular pentagon, some similar triangles and the Pythagorean theorem...

Written Article:

Interesting article about Gauss-Wantzel Theorem:

[hamsterdam]

I know I'm being pedantic, but any well-defined function has an exact value at every point. The phrase you're looking for is "expressible in terms of radicals" (without involving complex numbers)

[austin]

Not only is sin(pi/7) algebraic (as others have pointed out) but it has an “exact value” (by which I mean you can write it in terms of whole numbers, addition, subtraction, multiplication, division and root extraction - the technical term for such a number is a “solvable number”).

It’s minimal polynomial has degree 6, but all the exponents are even, which means it’s a cubic polynomial in x^2. Cubic polynomials are all solvable (there’s a cubic formula), and so the six roots of the minimal polynomial are obtained by taking the (two) square roots of each of the three solutions of the cubic.

What is true about sin(pi/7) is that it is not constructible (which means given a line segment in the plane of length 1, you cannot construct a line segment of length sin(pi/7) with just a straightedge and compass). This is because constructible numbers are precisely the numbers that can be written as expressions involving whole numbers, +, -, *, /, and SQUARE roots (as opposed to arbitrary nth roots, which is what distinguishes the definition of a solvable number from a constructable number). Indeed, the expression for sin(pi/7) involves taking cube roots.

[wqltr]

I swear sinpi/7 is algebraic but non-constructable

[Tetraverse]

sin(pi/7) is one of the roots of 64x^6-112x^4+56x^2-7

[alonamaloh]

sin(pi/7) is certainly algebraic. The degree of its minimal polynomial is 6, which makes it not constructible (because a constructible number would have minimal polynomial whose degree is a power of 2).

[mathmachine]

2:52 yes it is. It's a root of 64x⁶-112x⁴+56x²-7.

What it isn't is constructible. It cannot be expressed using only integers, addition, subtraction, multiplication, division, and square roots.

Also, adding cube roots into the mix doesn't fix this problem, unless you introduce complex numbers. Technically, though, there aren't ANY trig functions of rational multiples of π which arent constructible but are representable using cube roots of reals (though I'd love to be proven wrong). Same goes for fifth roots, sixth roots, seventh roots, etc. Only square roots and 2^n-th roots.

[Gyan-fxzx]

Sin(pi/7) is one root of
64x^6-112x^4+56x^2-7=0
Take x² = y
64y^3-112y^2+56y-7=0
So sin(pi/7) is a root of a polynomial equation as well as it can be represented in terms of radicals as we have the formula for a cubic. I got the above equation by expanding sin(7x) in terms of sin(x).

[PianoTriceratops]

sin(pi/7) is Algebraic; it's the imaginary part of the first root of the polynomial x^6 + x^5 + x^4 + x^3 + x^2 + x + 1. But since it requires cube roots to express, it is not Constructible. Constructible numbers are numbers that can be written using only a finite number of addition, multiplication, or square root operations. Otherwise nice video and beautiful geometric demonstration for the Golden Triangle.

[andrewwestcott]

The question that comes to mind is - is there an exact solution for ANY 'exact' value of sin, i.e. if the input can be written in some finite way, is there an output that can be written in some finite way (even if it involves millions of terms)

[brandonklein]

Respectfully i feel like you're glossing over the most important point at 1:26, the equality of the values of the sides of the rightmost isosceles triangle to the radial segment is not obvious but is the key insight that let's you apply simple geometric ideas to find sin(π/5).

[vishalmishra]

*Simplest derivation as a positive root of a quadratic equation*
5 x 18° = 90°, therefore 2 sin 18° cos 18° = sin (2x18° = 36°) = cos (3x18° = 54°) = 4 cos^3 18° - 3 cos 18° = cos 18° (4 cos^2 18° - 3 = 1 - 4 sin^2 18°)
Canceling cos 18° from both sides, we get the quadratic 4 x^2 + 2 x - 1 = 0, where x = sin 18° (positive root of the quadratic) and - sin 54° (negative root of the quadratic, since 5 x -54° = -270° same as +90°)
Therefore, cos 72° = sin 18° = (√5 - 1) / 4 and cos 36° = sin 54° = (√5 + 1) / 4
Hence, sin (π/5) = sin (180°/5) = sin 36° = √(1 - cos^2 36°) = (10 - √20) / 4.
Therefore, sin (18° N) and cos (18° N) are computable for any integer N, positive, zero or negative N.

[nicolastorres]

sin(pi/9) also has an expression with roots but it has to use complex numbers

[dylanm.]

Is it just me, or is 7 is just such a pain in the neck in everything ever? I hate 7. I even remember thinking about this as a child, just looking at a foam cutout of that wretched integer and thinking "man, this thing sucks".

I don't care if it's mathematically invalid. 7 = 0/10.

[ProactiveYellow]

Shouldn't sin(π/7) be the imaginary portion of one of the roots of x⁷+1=0? Given that it factors into (x+1)(x⁶-x⁵+x⁴-x³+x²-x+1)=0, it should be in the factors of that hexic factor. The fundamental theorem of algebra states that any polynomial with real coefficients can be factored into linear or quadratic terms with real coefficients, and since we know that the only real solution is x=-1, this hexic should be a triplet of quadratic factors. I guess my question is how an algebraic number can have a non-algebraic component.

[batchrocketproject]

This derivation depends entirely on the assumption (at 1:29) that the triangle can be divided to become golden with chord 'a' equal to the short legs on the left. Without clear reason to suppose that geometry holds, the derivation is circular - generating an answer that is consistent with the assumption made. One way to prove the triangle at 1:29 is golden, would be to work backwards from the value of sin pi/5 (hence circularity). It is not enough to claim the triangle to be "a known property of a regular pentagon", because the same argument can be made for any features (such as a value for sin pi/5).

[Geek]

I think why π/5 is largely skipped in secondary and university level math courses is basically due to utility. Much of the practical world uses triangles, squares and hexagons in some form or another. I wonder if easy tesselation of the plane partially explains that bias. Try finding a pentagon in everyday reality and it’s rather difficult. The most obvious example is the Pentagon outside Washington DC but there’s not much beyond that. You could make the argument that five-pointed stars are just stellated pentagons but the outlay of the shape is missing and doesn’t lead to an easily visible analysis on site. I doubt anyone looks at stars on most flags and say “there’s a 36° degree angle” or “there’s a pentagon hidden in plain sight”.

[jakobr_]

cool video but these sound effects are *really* irritating at times

[shruggzdastr-facedclown]

Numberphile has a great video on the constructability of heptadecagons

[dalibormaksimovic]

Indeed sin1 in degrees has exact value. It can be derived from cubic equation involving sin3. And sin 3 = sin (18-15). sin 18 is derived from half angle formula of sin 36, and sin15 is half angle formula for 30

[christianurso]

cos(pi/5) is phi/2. Doing a math test i found this out by making the arccosine of phi/2 and it turned out to be precisely pi/5. So i proved it with a pentagon. Now i can also say that sin(2pi/5) is also phi/2. Then i found out someone else also proved it in a similar way on the internet.