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The arc length of one sine period is 42E(1/2)

Example

For 0<k<1, write

E(k):=0π/21k2sin2tdt

for the complete elliptic integral of the second kind. The graph of sine over one period, parametrized by γ(x)=(x,sinx) for 0x2π, has arc length

L(γ)=42E(1/2).

Facts & Assumptions

Given: The graph path γ(x)=(x,sinx) on [0,2π].

[L1]
[L2]

(sinx)=cosx on R (The derivatives of sine and cosine are cosine and minus sine).

[L3]

The supplementary, reflection, and quarter-turn identities give cos(πx)=cosx, cos(π/2x)=sinx, and cos(x)=cosx (Cofunction, supplementary, quarter-turn, and reflection identities for the six trigonometric functions).

[L6]

Every positive real has a unique positive square root (Existence and uniqueness of n-th roots: a unique a1/n0 with (a1/n)n=a, case n=2).

[L7]

Differentiable real functions are continuous (A function differentiable at c is continuous at c).

[L8]

The number π=2γ is positive because the smallest positive zero of cosine satisfies γ(0,2) (Pi as twice the smallest positive zero of cosine, Cosine has a smallest positive zero, lying strictly between zero and two).

[L9]

For αR, the function uuα is continuous on (0,); in particular the positive square-root function is continuous there (Continuity and derivatives of positive-base real powers).

Verification

technique · direct
1.1

If 0<k<1, then [L5] gives 1k2sin2t1k2>0. By [L2], [L7], [L9], and [L10], the integrand defining E(k) is continuous on [0,π/2] and therefore integrable. Thus E(k) is well defined on the stated range.

L2L5L7L8L9L10algebra
1.2

The derivative in [L2] is continuous by [L7], and [L8] gives 0<2π, so [L1] gives L(γ)=02π1+cos2xdx.

L1L2L7L8
2.1

Split the integral in step 1.2 at π/2, π, and 3π/2. The affine reflections and translations licensed by [L4], together with [L3] and the square on cosine, show that all four pieces equal 0π/21+cos2xdx.

step 1.2L3L4
3.1

The modulus k:=1/2 exists by [L6] and satisfies 0<k<1. By [L5], 1+cos2x=2sin2x=2(112sin2x), so the quarter-period integral in step 2.1 is 2E(1/2).

step 2.1L5L6constructalgebra
4.1

Multiplying the quarter-period value in step 3.1 by the symmetry factor in step 2.1 yields L(γ)=42E(1/2).

step 2.1step 3.1algebra

Depends on

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Sources