Since the sine and cosine functions repeat its values over an interval of length `2pi` , hence the period of these trigonometric functions is `T = 2pi` , such that:

`f(x + 2pi) = f(x)`

Considering the sine function, `f(x) = sin x` , you may prove that `f(x + 2pi) = f(x)` , such that:

`f(x + 2pi) = sin (x + 2pi) => f(x + 2pi) = sin x*cos 2pi + sin 2pi*cos x`

Since `cos 2pi = 1` and `sin 2pi = 0` yields:

`f(x + 2pi) = sin x = f(x)`

Hence, evaluating `f(x + 2pi)` yields an equal value to the value of `f(x)` .

Considering the cosine function, `f(x) = cos x,` you may prove that `f(x + 2pi) = f(x)` , such that:

`f(x + 2pi) = cos (x + 2pi) => f(x + 2pi) = cos x*cos 2pi - sin 2pi*sin x`

`f(x + 2pi) = cos x = f(x)`

Hence, evaluating `f(x + 2pi)` yields an equal value to the value of `f(x)` .

**Hence, evaluating the period of both trigonometric functions sine and cosine, yields **`T = 2pi.`