IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

hsbhatt

We want to prove that sin^2kx has a period

Let the period of f(x) = sin^2kx be T

Then sin^2kx= sin^2k(x+T)

Let x+T = a.

So sin^2kx= sin^2ka

or sin^2kx- sin^2ka = 0.

We can now factorise using the identity xⁿ-yⁿ = (x-y) (x^n-1+x^n-2y+...+y^n-1)

Hence (sin²x - sin²a) [(sin²x)^k-1+(sin²x)^k-2(sin²a)+....+ [(sin²a)^k-1] = 0

The second factor is non-zero in the general case.

So sin²x = sin²a or cos2x = cos2a

Hence 2(x+T) = 2x+2n

or T = n

Thus

is the period in this case.