Finding range of $\sin^{20}(\theta)+\cos^{30}(\theta)$. We have to find the range of 
$$\sin^{20}(\theta)+\cos^{30}(\theta)$$
I have found the upper limit which is $1$.
I am a high school student and we were taught to convert functions into a simpler function to calculate the range.
I would like to know what kind of method is used in these types of problems. 

Here is a graph to clear some confusions in comments
 A: Here is an idea:
$$\sin^{20}\theta +\cos^{30}\theta$$
$$\sin^{20}\theta +(1-\sin^2\theta)^{15}$$
Put $\sin^2\theta =t$. Notice that $t \in [0,1]$ for no restriction on the domain.
Now $$f(t) = t^{10} +(1-t)^{15} \ \ \forall \ \ t \in [0,1] $$
$f(x)$ is continuous, so the extremum will occur at boundary points or where the derivative is $0$.
$$f(1) = 1$$
$$f(0) = 1$$
$$f'(t) = 10t^9-15(1-t)^{14}$$
Now you just wish you have a calculator or Wolfram Alpha to calculate the zero of this .
It comes out to be : $$t=0.43289180011435746401...$$
The value of $f(t)$ at this $t$ is $\approx 0.000432928$
So the range is :
$$f(t) \in [\approx 0.000432928, 1] $$
Wolfram Alpha link to calculation of point of Minima.
I noticed someone said that the lower bound tends to $0$. This is clearly not the case.
Zoomed-in image of the graph in the question at point of minima (It clearly does not approach zero.)

A: You can calculate the derivative of this function, and look for where it is zero. This will correspond to the local extremes.
