Odd Angles Formula Trivia

[Albert Chan]

(12-15-2019 01:49 AM)Albert Chan Wrote:  2⁵ sin(x)⁶ = -cos(6x) + 6 cos(4x) - 15 cos(2x) + 20/2

Since even powers of sines expanded to sum of cosines + constant, we have:

Quote:\(\large \int _{-\pi \over 2} ^{\pi \over 2} (\sin x)^{2n} \;dx = \binom{2n}{n}\pi / 2^{2n} \)

Example, for perimeter of ellipse, where h = (a-b)/(a+b)

\(p = (a+b) \int _{-\pi \over 2} ^{\pi \over 2} \sqrt{(1 + h^2)+ 2h \sin x} \;dx \)

Let \(y={h \over 1+h^2}\;, z = 2y \sin x\), and expand the integrand:

\(\sqrt{1 + h^2 + 2h \sin x} = \sqrt{(1+h^2)(1+z)} = \sqrt{1+h^2} \left[1 +
\binom{½}{1}z + \binom{½}{2}z^2 + \binom{½}{3}z^3 + \binom{½}{4}z^4
+ \binom{½}{5}z^5 + \binom{½}{6}z^6 +\;...\right ]\)

We can remove odd powers of z, since the area cancelled out.
Applying even powers of sine integral formula, we have:

\(p = \pi (a+b) \sqrt{1+h^2} \left[1 + \binom{½}{2}\binom{2}{1}y^2
+ \binom{½}{4}\binom{4}{2}y^4 + \binom{½}{6}\binom{6}{3}y^6 +\; ... \right]
= 2 \pi \sqrt{{a^2+b^2 \over 2}} \left[ 1 - {1\over 4}y^2 - {15\over 64}y^4
- {105 \over 256} y^6 -\; ... \right] \)

Unfortunately, convergence is much slower than series expansion on h.
However, we just proved \( p ≤ 2 \pi \sqrt{{a^2+b^2 \over 2}} \)

For a rough estimate, we can use Trapezoid's rule (2 trapezoids).

\( p ≈ \pi (a+b) \left({(1-h)\;+\; 2\sqrt{1+h^2}\;+\;(1+h) \over 4}\;\right)
= \pi (a + b) \left({1 + \sqrt{1+h^2} \over 2}\;\right)
= \pi \large\left( {a+b \over 2} + \sqrt{{a^2+b^2 \over 2}}
\;\right)\)