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问题: 三角函数

解答:

(1)
f(x) = √3sinωx-2sin²(ωx/2)+m
   = √3sinωx+cosωx + m-1
   = 2sin(ωx+π/6) + m-1
最小正周期=2π/ω=3π--->ω=2/3
x∈[0,π]--->ωx+π/6∈[π/6,5π/6]
    --->最小值=f(0)=2sin(π/6) + m-1 = 0--->m=0
--->f(x) = 2sin(2x/3+π/6) - 1

(2)f(C)=2sin(2C/3+π/6)-1=1
--->2C/3+π/6=π/2--->C=π/2=A+B
--->sin²B=cosB+cos(A-π/2)--->cos²A=2sinA
--->sin²A+2sinA-1=0--->sinA=√2-1