# Given `f(x) = sin^n x` , `x_n` coordinate inflecting point `0 <x_n<pi/2` , prove `sin x_n=sqrt((n-1)/n)` in `n>= 3` ?

*print*Print*list*Cite

### 1 Answer

Since the problem provides the information that `x_n` is the x coordinate of inflection point, hence, you need to evaluate the second order derivative such that:

`f'(x) = n*sin^(n-1) x*cos x`

You need to evaluate the second order derivative using the product rule such that:

`f''(x) = n*(n-1)*sin^(n-2) x*cos^2 x + n*sin^(n-1)x*(- sin x)`

`f''(x) = n*(n-1)*sin^(n-2) x*cos^2 x- n*sin^n x`

Factoring out `n*sin^(n-2) x` yields:

`f''(x) = n*sin^(n-2) x((n - 1)cos^2 x - sin^2 x)`

Since the x coordinate of an inflection point is the root of the second order derivative, hence `f''(x_n) = 0` such that:

`n*sin^(n-2) x_n((n - 1)cos^2 x_n - sin^2 x_n)`

Since the problem provides the information that `x_n in (0,pi/2), ` hence `sin x_n in (0,1), ` thus `sin^(n-2) x_n != 0` and `(n - 1)cos^2 x_n - sin^2 x_n = 0` such that:

`(n - 1)cos^2 x_n - sin^2 x_n = 0`

You need to use the fundamental formula of trigonometry such that:

`cos^2 x_n = 1 - sin^2 x_n`

`(n - 1)(1 - sin^2 x_n) - sin^2 x_n = 0`

You need to open the brackets such that:

`n - nsin^2 x_n - 1 + sin^2 x_n - sin^2 x_n = 0`

Reducing like terms yields:

`n - nsin^2 x_n - 1 = 0 => n - nsin^2 x_n = 1`

You need to isolate the term that contains `x_n` to the left side such that:

`-nsin^2 x_n = 1 - n => nsin^2 x_n = n- 1`

`sin^2 x_n = (n - 1)/n => sin x_n = +-sqrt((n - 1)/n)`

You need to keep only the positive value since for `x_n in (0,pi/2), ` sin `x_n in (0,1)` such that:

`sin x_n = +sqrt((n - 1)/n)`

**Hence, evaluating `sin x_n` for `x_n in (0,pi/2)` and `x_n` the `x` coordinate of the inflection point, yields `sin x_n = +sqrt((n - 1)/n), ` hence, the given identity holds.**