# Maximum and Minimum Values using Trigonometric Functions: $f(t)=t-2\cos(t)$ for $t\in[-\pi, \pi]$

RESOLVED

How would I go about finding the maximum and minimum values for an expression I cannot factor to find zeros?

I have some function:

$$f(t)=t-2\cos(t)$$

With the interval $[-\pi, \pi]$.

I would like to know how to find the absolute maximum and absolute minimum values specifically.

## 2 Answers

Based on the comments and your answer above, I assume you understand the process of finding absolute extreme and are just having trouble with finding the critical values. We found the derivative to be $f(t)=1+2sin(t)$, and by definition of critical numbers we want to see when this is equal to zero. Rearranging, we have $sin(t)=-1/2$, and now we have to use our knowledge of the unit circle. We are really trying to think backwards, and ask ourselves the sin of what, or in other words $t=arcsin(-1/2)$. This gives us $t=-5\pi/6$, and $t=\pi/6$. From now follow tips given by Mark to find the extreme points.

• Thank you. This is the part I wanted answered most. Very great explanation! – Computer May 30 '17 at 7:11
• Only I understand that $\sin^{-1}(-1/2)$ is equivalent to $-\pi/6$... where does the $-5\pi/6$ come into play? – Computer May 30 '17 at 7:14
• If you imagine the unit circle, $-\pi/6$ and $-5\pi/6$ are both the same vertical distance from the x-axis, just reflected along the y-axis, so they both have a sin value of -1/2. In fact, since all sin functions are periodic, we can have $-\pi/6 +k$, where $k$ is any integer, and it would have a sin value equal to -1/2. However we can neglect those solutions since we only look in $[0,2\pi]$. – Tyler6 May 30 '17 at 7:19
• Oh okay. Thank you very much! – Computer May 30 '17 at 7:21
• Happy to help :-) – Tyler6 May 30 '17 at 7:21

HINT:

Note that

$$f'(t)=1+2\sin(t)=0$$

when $t=-\pi/6$ and $t=-5\pi/6$.

At which of these values of $t$ is $f(t)$ a local maximum?

What are the values of $f(t)$ at the end points $-\pi$ and $\pi$?

• $-\pi/6$... But I do not know how you derived these numbers – Computer May 30 '17 at 3:39
• $\sin(t)=-1/2$ when $t=-\pi/6$ and $t=-5\pi//6$. – Mark Viola May 30 '17 at 3:40
• And how do I know that $\sin(t)$ is equal to $-1/2$ at $-\pi/6$ and $-5\pi/6$? – Computer May 30 '17 at 3:44
• I forgot to mention these should be absolute maximums not local maximums/minimums – Computer May 30 '17 at 3:46
• @Computer how I am supposed to answer that $f(-\pi)=-\pi-2\cos(-\pi)=-\pi+2\,$, $f(\pi)=\pi-2 \cos(\pi)=\pi+2\,$. – dxiv May 30 '17 at 5:31