若tanθ =根号3/3,则cosa2-sin(a+π/6)cosa 的值是

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  • cos²a-sin(a+π/6)cosa

    =cos²a-(sinacosπ/6+cosasinπ/6)cosa

    =cos²a-(√3/2sina+1/2cosa)cosa

    =cos²a-√3/2sinacosa-1/2cos²a

    =1/2cos²a-√3/2sinacosa

    =1/2(cos²a-√3sinacosa)/1

    =1/2(cos²a-√3sinacosa)/(sin²a+cos²a) 分子分母同时除以cos²a

    =1/2(cos²a/cos²a-√3sinacosa/cos²a)/(sin²a/cos²a+cos²a/cos²a)

    =1/2(1-√3tana)/(tan²a+1)

    =1/2(1-√3*√3/3)/[(√3)²+1]

    =0