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March 25, 2017

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If α and β are two angles in Quadrant II such that tan α= -1/2 and tan β = -2/3, find cos(α+β)

Work:
cos(α+β) = [ 1 - (tan α)(tan β) ] / [ 1 + (tan α)(tan β)]

cos(α+β) = [ 1 - (-1/2)(-2/3) ] / [ 1 + (-1/2)(-2/3)]

cos(α+β) = [ 1 - 1/3 ] / [ 1 + 1/3 ]

cos(α+β) = [ 2/3 ] / [ 4/3 ]

cos(α+β) = 1/2

is this right?

  • Math - ,

    Unfortunately it doesn't look right, ad... from the very start.

    Could you rechceck your source from where you got the expression?

    [ 1 - (tan α)(tan β) ] / [ 1 + (tan α)(tan β)]
    represents, according to me,
    cos(α+β)/cos(α-β).

    All is not lost, though.

    From

    1/cos2(x)
    =sec2(x)
    =1+tan2(x)

    tan(A+B)=(tan(A)+tan(B))/(1-tan(A)tan(B))

    You have ample ammunition to get the required answer.

    Post your results for a check-up if you wish, or you can substitute arbitrary numbers for tan(α) and tan(β) to check by yourself.

  • Math - ,

    ohh okay thanks!

  • Math - ,

    tan^-1(1/2)=x=.463648
    tan^-1(2/3)=y.588003
    x+y+pi/2 = -z=2.62245
    z=-2.62245
    cos (z) = -.868243

    and

    tan a= -1/2
    tan b= -2/3
    tan (a+b)=(tan a +tan b)/(1-tan a.tan b)
    so tan (a+b)= -7/4
    sec^2 x-tan^2 x=1
    so
    sec (a+b)=√(1+49/16)
    =√(65/16)
    (a+b)=1/sec (a+b)
    so cos(a+b) is positive
    so cos(a+b)=.48˜.5

    ???

  • Math - ,

    tan^-1(1/2)=x=.463648
    α=tan-1(-1/2)=π-x=2.67795

    tan^-1(2/3)=y.588003
    β=tan-1(-2/3)=π-y=2.55359

    α+β = 2.67795+2.55359 = 5.23154 rad. (serves as a check)
    cos(α+β) = 0.49614 (as a check)


    and

    tan a= -1/2
    tan b= -2/3
    tan (a+b)=(tan a +tan b)/(1-tan a.tan b)
    so tan (a+b)= -7/4
    sec^2 x-tan^2 x=1
    so
    sec (a+b)=√(1+49/16)
    =√(65/16)
    Therefore
    cos(α+β)
    =√(16/65)
    =4/√65

    since α+β is in the 4th quadrant, cos(α+β)>0

    Therefore
    cos(α+β) = +4/√65

  • Math - ,

    oh! thanks for your time!

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