已知函数f(x)=(1+cotx)sin²x+msin(x+π/4)× sin(x-π/4)
(1)当m=0时,求f(x)在区间[π/8,3π/4]上的取值范围(2)当tanα=2时,f(α)=3/5,求m的值...
(1)当m=0时,求f(x)在区间[ π/8 , 3π/4 ]上的取值范围
(2)当tanα=2时,f(α)=3/5,求m的值 展开
(2)当tanα=2时,f(α)=3/5,求m的值 展开
1个回答
展开全部
(1)m=0时,f(x)=(1+cotx)sin²x=1+sinx*cosx=0.5sin(2x)+1
在区间[ π/8 , 3π/4 ]上的取值范围为[0.5 , 1.5 ]
(2)当tanα=2时,cotα=1/2, sin²α=4/5
3/5=f(α)=(1+1/2)(4/5)+m(-1/2)(cos(α+π/4-α+π/4)-cos(α+π/4+α-π/4))
=6/5+ (-m/2)(1-cos2α)
=6/5+ (-m/2)(1-1+2sin²α)
=6/5+ (-m/2)(2*4/5)
m=3/4
在区间[ π/8 , 3π/4 ]上的取值范围为[0.5 , 1.5 ]
(2)当tanα=2时,cotα=1/2, sin²α=4/5
3/5=f(α)=(1+1/2)(4/5)+m(-1/2)(cos(α+π/4-α+π/4)-cos(α+π/4+α-π/4))
=6/5+ (-m/2)(1-cos2α)
=6/5+ (-m/2)(1-1+2sin²α)
=6/5+ (-m/2)(2*4/5)
m=3/4
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询