Thread: Extrema on a closed interval

Hi. This calculus problem has given me fits for the last few days:

Find the extrema of y=sinx + cosx on the interval [0,2pi]

I tried going at it regularly by taking the first derivative (y'=cosx - sinx) then setting it equal to 0. I just can't solve for "x" past that point. I think logically I should do (cosx = sinx), but then I can't remember that ancient trig. stuff to solve for "x". I could be going at it all wrong, though.

Does anyone have any suggestions?

This can seem a bit tricky at first trying to remember when sin(x)=cos(x). Think of your special triangles. Remember the 45-45-90 one? It has legs of 1 and 1 and a hypotenuse of square root of 2. The sine and cosine are the same here. So at 45 deg. or , sin(x)=cos(x). But like most trig functions, it's going to repeat itself with more than one angle. What other angles in the domain is this true?

You can also think of it another way using that

EDIT: Hint - One of the main trig functions is negative in the 2nd, 3rd, and 4th quadrant. How does restrict sine and cosine being equal, meaning which quadrants require sine and cosine to be opposite signs?

The other place that it repeats is 5pi/4, right?


So if the critical numbers are pi/4 and 5pi/4, then the extrema occur at x=sqrt(2) and x=-sqrt(2) ?

EDIT - The 2nd and 4th quadrant require the signs of sin and cos to be opposites. Does that have anything to do with the answer or was it just to help?

Yes, looks good to me. Now you have to plug in those values and decide whether the relative extrema are larger or smaller than than your function on the end points of your restricted domain.

Plug them back in to the original equation? That would make sense, so that would mean:

f(sqrt(2)) = 1.143...
f(-sqrt(2)) = -0.8318...
2pi = 6.2831
0 = 0

So there is a local minimum at x = -sqrt(2) and a local maximum at 2pi

Assuming your numbers above are correct, your reasoning using them is correct.