# 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 ?

sciencesolve | Certified Educator

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.