这几道数学题这么求啊(1)若cosα=1/7,α∈(0,π/2),则cos(α+π/3)=(2)化简cos(π/3+α)

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  • (1)先求sina取正的再代入,自己带吧

    (2)化简cos(π/3+α)+sin(π/6+α)=cos(π/3+α)+cos[π/2-(π/6+α)]

    =cos(π/3+α)+cos(π/3-α)

    =cosπ/3casα-sinαsinπ/3+cosπ/3casα+sinαsinπ/3

    =2cosπ/3casα=casα

    (3)[(1+cos20°)/2sin20°]-sin10°(1/tan5°-tan5°)

    =[(1+cos20°)/2sin20°]-sin10°(cot5°-tan5°)

    =[(1+cos20°)/4sin10°cos10°]-sin10°(cot5°-tan5°)

    =(2cos10°/4sin10°)-2sin5°cos5°(cot5°-tan5°)

    =(cos10°/2sin10°)-2((cos5°)^2-(sin5°)^2)

    =(cos10°/2sin10°)-2cos10°

    =(cos10°-4sin10°cos10°)/2sin10°

    =(sin80°-2sin20°)/2sin10°

    =((sin80°-sin20°)-sin20°)/2sin10°

    =(2cos50°sin30°-sin20°)/2sin10°

    =(sin40°-sin20°)/2sin10°

    ={2cos30°sin10°)/2sin10°

    =cos30°

    =(根号3)/2

    (4)tan2α直接用公式,sina+根3分之一,cosa=根下(2/3)带入求sin(2α+π/3)

    (5)sin163sin223+sin253sin313

    =sin(180-17)sin(180+43)+sin(180+73)sin(360-47)

    =-sin17sin43+sin73sin47

    =-cos73sin43+sin73cos43

    =sin(73-43)

    =sin30

    =1/2

    (6)tan(α+π/4)=1/7展开(tana+1)/(1-tana))=1/7

    解tana=-3/4推出sina=3/5,cosa=-4/5