Given f(x)=sin(x/2), show f(x+4pie)=f(x)?

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You need to evaluate `f(x + 4pi)` to verify if the equation of the function `f(x)` changes, hence, you need to replace `x + 4pi` for `x` in equation of the function, such that:

`f(x + 4pi) = sin((x + 4pi)/2)`

`f(x + 4pi) = sin(x/2 + (4pi)/2)`

`f(x + 4pi) = sin(x/2 + 2pi)`

If you do not remember that the principal period of sine function is `T = 2pi` , you may expand `sin(x/2 + 2pi)` , using the following trigonometric identity, such that:

`sin(a + b) = sin a*cos b + sin b*cos a`

Reasoning by analogy, yields:

`sin(x/2 + 2pi) = sin(x/2)*cos(2pi) + sin(2pi)*cos(x/2)`

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

`sin(x/2 + 2pi) = sin(x/2) => f(x + 4pi) = sin(x/2) = f(x)`

**Hence, evaluating `f(x + 4pi)` yields that the equation of the function `f(x)` does not change, thus the statement `f(x + 4pi) = f(x)` holds.**

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