There is an alternative to justaguide's solution using infinite series. Also, the solution above is assuming `y=cosx` when evaluating at `x=+-pi/2` and `x=1` .

This function can be rewritten in a much simpler form since it is a geometric series with initial value 1 and common ratio `r=sinx` . Since the function `-1<sinx<1` on the domain `(-pi/2,pi/2)` , the function converges to `f(x)=1/{1-sinx}` since the common ratio `|r|<1` . Furthermore, we see that when `x=-pi/2` , the series is alternating and converges to 1. Also, when `x=pi/2` , the series does not converge, and approaches `infty` .

The maximum value of the series is then `infty` . The minimum value can be found by differentiating the function to get:

`f'(x)=cosx/(1-sinx)^2`

Setting it to zero and solving means that the minimum is at `cosx=0`

This happens when `x=-pi/2` . The solution at `x=pi/2` is the maximum.

**The range over the specified domain is `R=[1,infty)` . A graph of the function is given by:**

The range of a function f(x) is the set of values f(x) lies in for x lying in the domain.

Here, `f(x) = 1 + sin x + sin^2x + sin^3x ...` and the domain is `[-pi/2, pi/2]`

At x = `-pi/2` and `x = pi/2` , `sin x = 0`

At x = 0, sin x = 1

This gives the minimum value of f(x) as 1 and the maximum value is infinity.

**The range of `f(x) = 1 + sin x + sin^2x + sin^3x ...` is `[1, oo)` if the domain is `[-pi/2, pi/2]` **

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