What are the absolute extrema of #y=cos^2 x - sin^2 x# on the interval [-2,2]?

Question

2246 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-20T19:12:31

#cos^2x-sin^2x = cos(2x)# which has a maximum value of #1# (at #x=0#) and a minimum value of #-1# (at #2x = pi# so #x = pi/2#)