A) Given function:
[tex]f(\theta)=\sin{\theta}+4\cos{\theta}[/tex]
to determine the critical points we need to differentiate [tex]f(\theta)[/tex] and equate it to zero. (the slopes of the curve at the critical point is always zero)
[tex]\dfrac{d}{d\theta}(f(\theta))=\dfrac{d}{d\theta}(\sin{\theta}+4\cos{\theta})[/tex]
[tex](f'(\theta)=\cos{\theta}-4\sin{\theta}[/tex]
now use [tex](f'(\theta)=0[/tex]
[tex]0=\cos{\theta}-4\sin{\theta}[/tex]
now solve for [tex]\theta[/tex] within the range [tex][-2\pi,2\pi][/tex]
[tex]4\sin{\theta}=\cos{\theta}[/tex]
[tex]\dfrac{\sin{\theta}}{\cos{\theta}}=\dfrac{1}{4}[/tex]
[tex]\tan{\theta}=\dfrac{1}{4}[/tex]
[tex]\theta=\arctan\left({\dfrac{1}{4}\right)}[/tex]
[tex]\alpha=0.2450[/tex]
as positive tan lies in the first and third quadrant of the unit-circle. our values within the interval [tex][-2\pi,2\pi][/tex] will be:
[tex]\begin{tabular}{ccccc}\theta=&-2\pi+\alpha,&-\pi+\alpha,&0+\alpha,&\pi+\alpha&\theta=&-2\pi+0.2450,&-\pi+0.2450,&0.2450,&\pi+0.2450&\theta=&-6.038,&-2.897,&0.2450,&3.38,\end{tabular}\\[/tex]
These are the critical points!
We can use these values to find the values of [tex]f(\theta)[/tex]
B) From the graph we can see that the first and third are maxima and second and fourth are minima.
C) Since all minimum points have the same y-coordinate, and all maximum points have the same y-coordinate. We can safely say that all points are local critical points in this function.
