高一三角恒等1) 证明 tan^2 x + 1/(tan^2 x) = [2(3+2cos4x)]/(1-cos4x)

2个回答

  • (1)

    证明:

    tan^2(x)+1/[tan^2(x)]

    =[sin^2(x)/cos^2(x)]+[cos^2(x)/sin^2(x)]

    ={[sin^4(x)+cos^4(x)]/[sin^2(x)cos^2(x)]

    由于:

    分子:sin^4(x)+cos^4(x)

    =[sin^2(x)]^2+[cos^2(x)]^2

    =[(1-cos2x)/2]^2+[(1+cos2x)/2]^2

    =[1/4+1/4cos^2(2x)-1/2cos2x]

    +[1/4+1/4cos^2(2x)+1/2cos2x]

    =1/2+1/2cos^2(2x)

    =1/2+1/2*[(1+cos4x)/2]

    =3/4+1/4*cos4x

    =(3+cos4x)/4

    分母:sin^2(x)cos^2(x)

    =[(1-cos2x)/2]*[(1+cos2x)/2]

    =[1-cos^2(2x)]/4

    ={1-[(1+cos4x)/2]}/4

    =(1-cos4x)/8

    则:分子/分母

    =[(3+cos4x)/4]/[(1-cos4x)/8]

    =[2(3+cos4x)]/(1-cos4x)

    即命题得证

    (2)

    (1+tan22)(1+tan23)(1+tan24)(1+tan25)

    =[(1+tan22)(1+tan23)][(1+tan24)(1+tan25)]

    =[1+tan22*tan23+(tan22+tan23)]*[1+tan24*tan25+(tan24+tan25)]

    由于

    tan45

    =tan(22+23)

    =(tan22+tan23)/(1-tan22*tan23)

    =1

    所以

    tan22+tan23

    =tan45*(1-tan22*tan23)

    =1-tan22*tan23

    则:

    (1+tan22)(1+tan23)(1+tan24)(1+tan25)

    =[1+tan22*tan23+(tan22+tan23)]*[1+tan24*tan25+(tan24+tan25)]

    =[1+tan22*tan23+(1-tan22*tan23)]*[1+tan24*tan25+(tan24+tan25)]

    =2+2tan24*tan25+2(tan24+tan25)