Question

Prove that 2pi is the minimum period of f(x) = cos (x)

Answer 1

let y = cosx
dy/dx = -sinx
at a max or min, dy/dx = 0
-sinx = 0
sinx = 0
looking at a standard sine curve
x = 0, π, 2π, 3π, ...

when x = 0 , f(0) = cos0 = 1
when x = π, f(π) = cosπ = -1
wehn x = 2π, f(2π) = +1
when x = 3π, f(3π) - -1
.....

the max of f(x) = 1, the min of f(x) = -1
and f(π)= -1, thus a minimum