由于函数f(x)=msinx+ncosx=
m2 +n2] sin(x+∅),且f(
π
4)是它的最大值,
∴[π/4]+∅=2kπ+[π/2],k∈z,∴∅=2kπ+[π/4],∴tan∅=[n/m]=1.
∴f(x)=
m2 +n2 sin(x+2kπ+[π/4])=
m2 +n2 sin(x+[π/4] ).
对于①,由于 f(x+
π
4)=
m2 +n2 sin(x+[π/4]+[π/4] )=
由于函数f(x)=msinx+ncosx=
m2 +n2] sin(x+∅),且f(
π
4)是它的最大值,
∴[π/4]+∅=2kπ+[π/2],k∈z,∴∅=2kπ+[π/4],∴tan∅=[n/m]=1.
∴f(x)=
m2 +n2 sin(x+2kπ+[π/4])=
m2 +n2 sin(x+[π/4] ).
对于①,由于 f(x+
π
4)=
m2 +n2 sin(x+[π/4]+[π/4] )=