# Homework Help: Find the max and min of trig function without calc or drawing graph

1. Aug 26, 2013

### needingtoknow

1. The problem statement, all variables and given/known data

y = (4/3)cos(x/2 + pi/6) - (2/3)
(0 <= x < 2pi)

Find the maximum and minimum values for the trigonometry function without calculus or drawing the graph.

Basically in the solution they apply the transformations to the domain given in order to find a new domain and range, so I attempted to do the same.

So first I tried to get the new domain, by applying changes to 0 <= x < 2pi

0 <= x < 2pi
pi/6 <= x < 2pi/2 + pi/6

pi/6 <= x/2 + pi/6 < 7pi/6 // until this step is correct, no difficulties here

Now from the new domain I have to find a new range so I find the cos of the x values of the new domain

cos of pi/6 is +(root3)/2

cos of 7pi/6 is +(root3)/2

This is where I am confused because the range is the same value

+(root3)/2 <= cosx <= +(root3)/2

However, if I had gotten this step right what I would have done next is apply the transformations from the given function that apply to the y values. Why is the earlier step wrong though? Can some please help?

2. Aug 26, 2013

### tia89

I am not sure to have understood what you did, but I advice you if you want to go on this way to give names to things (change of variable) so that you are prevented from mistakes... anyway I think you are complicating things...
Think to the shape of a cosine... you know exactly well the graph (don't need to draw it) and you know where maxima and minima are, as they are properties of the function itself, no need of calculus to know them. Now reason on the function you are given... the 4/3 in front simply amplifies the function without displacing the maxima and minima somewhere else... the 2/3 subtracted as well shifts down vertically the function but again does not displace maxima and minima... therefore you have enough to find the arguments of the cosine to be 0 or 3/2$\pi$ and the trick is done... just remember that they can be shifted by $2\pi$ to match the original domain...
I should think this is the easiest way...

3. Aug 26, 2013

### verty

$cos(\frac{7π}{6}) = -\frac{\sqrt{3}}{2}$, the cosine of a third-quadrant angle is negative. For what it's worth, I don't think this problem can be solved without drawing the graph or using calculus unless one is given that the minimum of cosine on the interval $[\frac{\pi}{6}, \frac{7\pi}{6}]$ = -1. How would one discover that otherwise?