1:求值:tan70°×cos10°(根号3×tan20°—1)

4个回答

  • tan70×cos10×(√3tan20-1)

    =tan70×cos10×(tan60×tan20-1)

    =tan70×cos10×[(sin60×sin20/cos60×cos20)-1]

    =tan70×cos10×(sin60×sin20-cos60×cos20)/(cos60×cos20)

    =tan70×cos10×[-cos(60+20)]/(cos60 ×cos20)

    =-tan70×cos10×cos80/(cos60×cos20)

    =-tan70×cos10×sin10/(cos60×cos20)

    =-(sin70/cos70)×(1/2)×sin20/(cos60×cos20)

    =-(cos20/sin20)×sin20/(2cos60×cos20)

    =-1/(2cos60)

    =-1

    cos a=1/7 且0<a<π/2故sin a>0且

    sin²a=1-cos²=1-1/49=48/49故sin a=4√3/7

    而cos(a-b)=cos a cos b+sin a sin b=1/7cos b+4√3/7sin b=13/14即

    cos b+4√3sin b=13/2 ①

    由0<a-b<π/2知 sin (a-b)>0 而sin²(a-b)=1-cos²(a-b)=1-169/196=27/196

    sin(a-b)=3√3/14=sin a cos b-cos a sin b=4√3/7cos b-1/7sin b即

    4√3cos b-sin b=3√3/2 两边同时乘以4√3,有

    48cos b-4√3sin b=18 ②

    两式相加,有

    49cos b=49/2则cos b=1/2 由0<b<π/2知b =arc cos 1/2=π/3

    f(x)=2cos2x+sin²x-4cosx =2cos(2π/3)+sin²(π/3)-4cos(π/3)

    =2*(-1/2)+3/4-4*1/2

    =-9/4

    f(x)=2cos2x+sin²x-4cosx=2(2cos²x-1)+(1-cos²x)-4cosx=3cos²x)-4cosx-1

    =3(cosx-2/3)²-7/3

    当cosx=1时,最小值为-2

    f(x)=6cos^2x-√3sin2x

    =3(1+cos2x)-√3sin2x

    =3+2√3(cosπ/6cos2x-sinπ/6sin2x)

    =3+2√3cos(2x+π/6)

    fmax(x)=3+2√3

    fmin(x)=3-2√3

    2)

    f(a)=fmin(x)

    2a+π/6=π,a=π/2-π/12=5π/12

    4a/5=π/3

    tan4/5a=tanπ/3=√3

    当cosx=-1时,最大值为6