# Show that the length of one arch of the sine curve is equal to the length of one arch of the cosine curve.

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`