on the integral

But how do I find out these two calculations below without using a calculator?

and

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- Mar 12th 2008, 06:40 PM!!![SOLVED] Determing absolute minimum and maximum values
on the integral

But how do I find out these two calculations below without using a calculator?

and

- Mar 12th 2008, 06:53 PMThePerfectHacker
The points you have to check to know which are largest and smallest are: . To determine which is largest perform the same operations on this triple. First multiple each by 2: . Now square each one: . Subtract 4 from each one: . And square again: . We see that that 1st point is the smallest, 3rd point is in between, and 2nd point is largest.

- Mar 12th 2008, 07:09 PM!!!
Ahh, thank you.

But how do I determine from and from ? - Mar 12th 2008, 07:40 PMThePerfectHacker
- Aug 7th 2014, 05:02 AMHallsofIvyRe: [SOLVED] Determing absolute minimum and maximum values
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 , they are each . The hypotenuse has, by the Pythagorean theorem, length so that equals "opposite side over hypotenuse" which is and the same for the cosine.

For and construct an equilateral triangle (all three side of equal length so all three angles congruent so each angle ) 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 , another angle , 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 angle and the perpendicular constructed) has length .

From that, we have , and , .