设f（x）=√sinx+2cos^2x，求f（π/2）f（-π/2）

更新时间：2021-10-13 10:07:30 70次访问

f(π/2) = √sin(π/2) + 2[ cos(π/2) ]^2

= √1 + 2[ 0 ]^2

= 1；

f(-π/2) = √sin(-π/2) + 2[ cos(-π/2) ]^2

= √(-1) + 2[ 0 ]^2

= i 。

f(-π/2) 在实数域不存在。

f(x)=√sinx+2cos²x

f(π/2)=√1+2*0²=1

f(-π/2)=√(-1)+2*0²=i