Question

已知函数f(x)=(1+cotx)sin^2x+msin(x+π/4)sin(x-π/4).当tanα=2时，f(α）=3/5，求m的值。

Answer 1

f(x)=(sinx)^2+sinxcosx+1/2*m(cosx)^2-1/2*m(sinx)^2
=(cosx)^2[(tanx)^2+tanx+1/2*m-1/2*m(tanx)^2]
tanα=2
(cosa)^2=1/5
f(α）=1/5*(6-3/2*m)=3/5
m=2

希望能帮到你，祝学习进步O(∩_∩)O,也别忘了采纳！

Answer 2

f(x)=(1+cotx)sin^2x+msin(x+π/4)sin(x-π/4).
=(1+cotx)sin^2x+m(√2/2sinx+√2/2cosx)(√2/2sinx-√2/2cosx)
=(1+cotx)sin^2x+m(1/2sin^2x-1/2cos^2x)
=[(1+cotx)tan^2xcos^2x+1/2m(tan^2xcos^2x-cos^2x)]/(sin^2x+cos^2x)
=[(1+cotx)tan^2x+1/2m(tan^2x-1)]/(tan^2x+1)
因为当tanα=2时，cotx=1/2，f(α）=3/5，
所以[(1+1/2)*2^2+1/2m(2^2-1)]/(2^2+1)=3/5，即（6+3/2m)/5=3/5，解得m=-2

Answer 3

cot(α)=1/2,
(cscα)^2=1+1/4=5/4,
(sinα)^2=4/5,
f(α)=(1+1/2)*(4/5)+m[(sinα)^2-(cosα)^2]/2
=6/5-(mcos2α)/2
=6/5-m*[1-2(sinα)^2]/2
=6/5-m(1-8/5)/2
=6/5+3m/10
6/5+3m/10=3/5,
m=-2.