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  • 已知tanθ=2√2,θ∈(π/4,π/2),求[2cos²θ/2-sinθ-1]/[√3sin(π/3+θ)sin(π/3-θ)]的值

    已知tanθ=2√2,θ∈(π/4,π/2),求[2cos²θ/2-sinθ-1]/[√3sin(π/3+θ)sin(π/3-θ)]的值
    主要是√3sin(π/3+θ)sin(π/3-θ)怎么化成√3/2cos2θ+√3/4
    數(shù)學(xué)人氣:871 ℃時(shí)間:2020-06-03 13:40:09
    優(yōu)質(zhì)解答
    你好
    √3sin(π/3+θ)sin(π/3-θ)=√3/2{cos[(π/3+θ)-(π/3-θ)]-cos[(π/3+θ)+(π/3-θ)]}
    =√3/2{cos[(π/3+θ-π/3+θ)]-cos[(π/3+θ+π/3-θ)]}
    =√3/2{cos(2θ)-cos[(2π/3)]}
    =√3/2[cos2θ-cos(2π/3)]
    =√3/2cos2θ-√3/2cos(2π/3)
    =√3/2cos2θ-√3/2cos(120°)
    =√3/2cos2θ-√3/2*(-1/2)
    =√3/2cos2θ+√3/4
    注意cos[(π/3+θ)-(π/3-θ)]-cos[(π/3+θ)+(π/3-θ)]
    =cos(π/3+θ)cos(π/3-θ)+sin(π/3+θ)sin(π/3-θ)-[os(π/3+θ)cos(π/3-θ)-in(π/3+θ)sin(π/3-θ)]
    =2sin(π/3+θ)sin(π/3-θ)
    即sin(π/3+θ)sin(π/3-θ)=1/2{cos[(π/3+θ)-(π/3-θ)]-cos[(π/3+θ)+(π/3-θ)]}請(qǐng)問(wèn)下√3sin(π/3+θ)sin(π/3-θ)=√3/2{cos[(π/3+θ)-(π/3-θ)]-cos[(π/3+θ)+(π/3-θ)]}這步使用的什么公式積化和差公式sinasinb=1/2[cos(a-b)-cos(a+b)]
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