Question

两道与三角函数有关的题目1.设a为常数,且a＞1,0≤x≤2π 则函数f(x)=cos^2X+2asinX-1的最大值为

Answer 1

(1)f(x)=cos^2X+2asinX-1

=-sinX^2+2asinX

=-(sinX-a)^2+a^2

0≤x≤2π ,-1≤sinX≤1

a＞1,sinX=1时,f(x)最大值为a^2

选D

（2）证明：sin(x+y)=1

x+y=π/2+2kπ

2x+y=π/2+2kπ+x

tan(2x+y)+tan y

= tan(π/2+2kπ+x)+tan(π/2+2kπ-x)

=tan(π/2+x)+tan(π/2-x)

=-cotx+cotx

=0