f(派/3)=1
1/2(√3a+b)=1
b=2-√3a
f(x)=asinx+bcosx
=√(a^2+b^2)sin(x+t),tant=b/a
=√(a^2+3a^2-4√3a+4)sin(x+t),
=2(√[(a-√3/2)^2+1/4]sin(x+t),
>=sin(x+t),a=√3/2等号成立
>=1(取得最大sin(x+t)=1)
f(x)>=1
f(派/3)=1
1/2(√3a+b)=1
b=2-√3a
f(x)=asinx+bcosx
=√(a^2+b^2)sin(x+t),tant=b/a
=√(a^2+3a^2-4√3a+4)sin(x+t),
=2(√[(a-√3/2)^2+1/4]sin(x+t),
>=sin(x+t),a=√3/2等号成立
>=1(取得最大sin(x+t)=1)
f(x)>=1