# Thread: [SOLVED] Determing absolute minimum and maximum values

1. ## [SOLVED] Determing absolute minimum and maximum values

$f(x)=sin(x)+cos(x)$ on the integral $[0, \frac{\pi}{3}]$
$f'(x) = cos(x) - sin(x) = 0$
$\frac{sin(x)}{cos(x)}=tan(x)=1=\frac{\pi}{4}$

$0, (\frac{\pi}{4}), (\frac{\pi}{3})$

$f(0)=sin(1)+cos(1)=0+1=1$
But how do I find out these two calculations below without using a calculator?
$f(\frac{\pi}{4}) = sin(\frac{\pi}{4}) + cos(\frac{\pi}{4})=\sqrt{2} = 1.44$
and
$f(\frac{\pi}{3}) = sin(\frac{\pi}{3}) + cos(\frac{\pi}{3}) = \sqrt{3} + 1 / 2 = 1.37$

2. Originally Posted by !!!
$f(x)=sin(x)+cos(x)$ on the integral $[0, \frac{\pi}{3}]$
$f'(x) = cos(x) - sin(x) = 0$
$\frac{sin(x)}{cos(x)}=tan(x)=1=\frac{\pi}{4}$

$0, (\frac{\pi}{4}), (\frac{\pi}{3})$

$f(0)=sin(1)+cos(1)=0+1=1$
But how do I find out these two calculations below without using a calculator?
$f(\frac{\pi}{4}) = sin(\frac{\pi}{4}) + cos(\frac{\pi}{4})=\sqrt{2} = 1.44$
and
$f(\frac{\pi}{3}) = sin(\frac{\pi}{3}) + cos(\frac{\pi}{3}) = \sqrt{3} + 1 / 2 = 1.37$
The points you have to check to know which are largest and smallest are: $1,\sqrt{2}, (\sqrt{3}+1)/2$. To determine which is largest perform the same operations on this triple. First multiple each by 2: $2,2\sqrt{2}, \sqrt{3}+1$. Now square each one: $4,8,4+\sqrt{3}$. Subtract 4 from each one: $0,4,\sqrt{3}$. And square again: $0,16,3$. We see that that 1st point is the smallest, 3rd point is in between, and 2nd point is largest.

3. Ahh, thank you.
But how do I determine $\sqrt{2}$ from $sin(\frac{\pi}{4}) + cos(\frac{\pi}{4})$ and $\sqrt{3} + 1 / 2$ from $sin(\frac{\pi}{3}) + cos(\frac{\pi}{3})$?

4. Originally Posted by !!!
Ahh, thank you.
But how do I determine $\sqrt{2}$ from $sin(\frac{\pi}{4}) + cos(\frac{\pi}{4})$ and $\sqrt{3} + 1 / 2$ from $sin(\frac{\pi}{3}) + cos(\frac{\pi}{3})$?
Because (you are expected to know) $\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{\sqrt{2}}2$. And $\cos \frac{\pi}{3} = \frac12$ and $\sin \frac{\pi}{3} = \frac{\sqrt{3}}2$

5. ## Re: [SOLVED] Determing absolute minimum and maximum values

Originally Posted by ThePerfectHacker
Because (you are expected to know) $\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{\sqrt{2}}2$. And $\cos \frac{\pi}{3} = \frac12$ and $\sin \frac{\pi}{3} = \frac{\sqrt{3}}2$
Construct a right triangle with legs each of length 1. Since this is an isosceles triangle, the two acute angles are congruent. Since they must add to $\pi/2$, they are each $\pi/4$. The hypotenuse has, by the Pythagorean theorem, length $\sqrt{2}$ so that $sin(\pi/4)$ equals "opposite side over hypotenuse" which is $\frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}$ and the same for the cosine.

For $\pi/6$ and $\pi/3$ construct an equilateral triangle (all three side of equal length so all three angles congruent so each angle $\frac{\pi}{3}$) with sides of length 2 and draw a line from one vertex perpendicular to the opposites side. By symmetry (that line divides the equilateral triangle into two congruent right triangles) we have a right triangle with one angle $(\pi/3)/2= \pi/6$, another angle $\pi/3$, hypotenuse of length 2 and side opposite the [tex]\pi/6[itex] angle of length 1. By the Pythagorean theorem the other leg (opposite the $\pi/3$ angle and the perpendicular constructed) has length $\sqrt{4- 1}= \sqrt{3}$.

From that, we have $sin(\pi/6)= \frac{1}{2}$, $cos(\pi/6)= \frac{\sqrt{3}}{2}$ and $sin(\pi/3)= \frac{\sqrt{3}}{2}$, $cos(\pi/3)= \frac{1}{2}$.