There are multiple ways to solve this problem but I will just compute the arc length of sine and cosine over half a period, or one hump. The arc length formula for a function is:

`L=int_a^b sqrt(1+f'(x)^2) dx`

Let f(x)=cos(x) and g(x)=sin(x). Then we want to see if:

`int_a^b sqrt(1+f'(x)^2) dx=int_c^d sqrt(1+g'(x)^2) dx`

`int_(-pi/2)^(pi/2) sqrt(1+cos(x)'^2) dx=int_0^pi sqrt(1+sin(x)'^2) dx`

`int_(-pi/2)^(pi/2) sqrt(1+sin(x)^2) dx=int_0^pi sqrt(1+cos(x)^2) dx`

`int_(-pi/2)^(pi/2) sqrt(2-cos(x)^2) dx=int_0^pi sqrt(2-sin(x)^2) dx`

`sqrt(2) int_(-pi/2)^(pi/2) sqrt(1-1/2 cos(x)^2) dx=sqrt(2) int_0^pi sqrt(1-1/2 sin(x)^2) dx`

Use symmetry to change the bounds of integration.

`2 sqrt(2) int_(0)^(pi/2) sqrt(1-1/2 cos(x)^2) dx=2 sqrt(2) int_0^(pi/2) sqrt(1-1/2 sin(x)^2) dx`

We need to manipulate the left hand side (LHS) to get it into a similar form as the right hand side. Make a dummie variable u-substitution and then use a trigonometric identity:`x=pi/2-u, dx=-du`

`LHS: 2 sqrt(2) int_(0)^(pi/2) sqrt(1-1/2 cos(pi/2-u)^2) (-du)= -2 sqrt(2) int_(0)^(pi/2) sqrt(1-1/2 sin(u)^2) du`

Change bounds of of integration to be in terms of u so we can drop the dummie variable.

`LHS: -2 sqrt(2) int_(pi/2)^(0) sqrt(1-1/2 sin(u)^2) du=2 sqrt(2) int_(0)^(pi/2) sqrt(1-1/2 sin(u)^2)`

Therefore,

`LHS=RHS`

`2 sqrt(2) int_(0)^(pi/2) sqrt(1-1/2 sin(u)^2)=2 sqrt(2) int_(0)^(pi/2) sqrt(1-1/2 sin(u)^2)`

This is an elliptic integral of the second kind with the form:

`E(phi,k)=int_0^phi sqrt(1-k^2 sin(theta)^2) d(theta), 0<k<1`

Where in this case `E(phi,k)=E(pi/2,1/2)`

This needs to be evaluated numerically but you will find that both sides are equal to:

`2 sqrt(2) E(pi/2,1/2)~~3.8202`

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