How do you sketch the curve #y=cos^2x-sin^2x# by finding local maximum, minimum, inflection points, asymptotes, and intercepts?

1 Answer
Jul 26, 2017

Note that we can use the trigonometric identity:

#cos^2x-sin^2x = cos 2x#

so we know that the function has a maximum for #x = kpi# and a minimum for #x= pi/2+kpi# with #k in ZZ#.

The inflection points are coincident with the intercepts and are in #x= pi/4+kpi#. The function is concave down in #(-pi/4+kpi, pi/4+kpi)# with #k# even, and concave up with #k# odd.

Finally, the function is continuous for every #x in RR# and has no limit for #x->+-oo# and no asymptotes.

graph{cos(2x) [-2.5, 2.5, -1.25, 1.25]}