# What is the period of the trigonometric function `f(x) = sin (sqrt(x))?`

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### 1 Answer

It doesn't have a period. To see this, note that `sin(0)=0` , and the first "cycle" of the function `sin(sqrt(x))` ends when `sqrt(x)=2pi` , or equivalently when `x=4pi^2.` If the function were periodic, then `4pi^2` would be the period. If you're working in degrees then it's the same idea, except instead of `2pi` radians we use `360^@.`

However, the second cycle is completed when you go twice around the unit circle, or when `sqrt(x)=4pi`, so `x=16pi^2.` The third cycle is completed when `sqrt(x)=6pi,` or `x=36pi^2.` Each cycle takes longer and longer to complete, so the graph of `y=sin(sqrt(x))` can't be formed by repeating one cycle. Here's the graph.

The reason is that unlike something like `sin((3x)/7),` which replaces the variable `x` with the *constant multiple *`3/7*x` , and thus horizontally stretches the graph *uniformly,* replacing `x` with `sqrt(x)` stretches the graph more and more as `x` increases, as you can see from the graph.

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