19. Using the double-angle expansion, we expand cos2x: cos2x=2cos? X- 1 (Of course, there is also the result of sine expansion, but in order to solve the problem conveniently, cos2x is expanded into cosine form).
So there is: y=2cosx-cos2x.
=2cosx-2cos？ x+ 1
=-2cos？ x+2cosx+ 1
The problem is transformed into the problem of solving the maximum value of a quadratic function with one variable by method of substitution.
Let cosx=t, t∈[- 1, 1]
Then there is: y=-2t? +2t+ 1
=-2(t？ -t)+ 1
=-2(t？ -t+ 1/4)+3/2
=-2(t- 1/2)？ +3/2
It is easy to know that when t= 1/2, that is, x=60 degrees, the maximum value obtained by the function is 3/2.
20. Using cosine theorem, BC*BC=AB*AB+AC*AC-2AB*AC*cosA, BC=7.
2 1, mean: (7+ 14+ 12+4+8)/5=9 variance: [(7-11) * (7-1/. ( 14- 1 1)+( 12- 1 1)*( 12- 1 1)+(4- 1 1)*(4- 1 1)+(8- / kloc-0/ 1)*(8- 1 1)]/5= 16.8