题目

y=根号6cosx-根号2sinx的值域

提问时间：2021-01-02

答案

y=√6cosx-√2sinx
=√2(√3cosx-sinx)
=2√2(√3/2*cosx-1/2*sinx)
=2√2(cosπ/6*cosx-sinπ/6*sinx)
=2√2cos(π/6+x)
-1<=cos(π/6+x)<=1
-2√2<=2√2cos(π/6+x)<=2√2
y∈[-2√2,2√2]