振幅A=√2
周期T=2*(6+2)=16=2π/ω
ω=π/8
2ω+φ=π/2+2kπ
φ=π/4+2kπ
f(x)=√2sin(πx/8+π/4)
f(x)在x∈[4,12]最大值是f(4)=1
∴c=4
由余弦定理:c²=a²+b²-2abcosC
1=a²+b²-ab≥2ab-ab
∴ab≤1
S=(absinC)/2≤√3/4
振幅A=√2
周期T=2*(6+2)=16=2π/ω
ω=π/8
2ω+φ=π/2+2kπ
φ=π/4+2kπ
f(x)=√2sin(πx/8+π/4)
f(x)在x∈[4,12]最大值是f(4)=1
∴c=4
由余弦定理:c²=a²+b²-2abcosC
1=a²+b²-ab≥2ab-ab
∴ab≤1
S=(absinC)/2≤√3/4