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Math Help - COS(A+B) Problem - Whats COS A?

  1. #1
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    COS(A+B) Problem - Whats COS A?

    Hi i'm new,

    I have been trying to solve this for a couple of days and had to just give up. Hopefully it is in the right topic.

    I have a system which gives me the answer to COS(A+B) and I can work out what B is.

    Therefore:-

    COS(A+B) = (Known Varible Output)
    B = (Known Constant)

    What is COS A ?
    (I don't need to know what A is - just COS A)

    The solution needs to work for a full COS cycle (i.e. 0 to 2PI) therefore the answer is not COS((COS-1(A+B))-B) !




    I have been trying to use COS(A+B)=COSACOSB-SINASINB
    and COS^2(A)+SIN^2(A)=1
    with no luck just can't seem to be able to rearange the formula.



    If anyone can solve this I would be very thankful/amazed.

    Dave
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: COS(A+B) Problem - Whats COS A?

    Given values for cos(A+B) and B, there are two possible values for cos(A). You're on the right track, all you need is to recognize that the arccosine of a number may be either positive or negative:

     \cos(A) = \cos(\pm \cos ^{-1}(\cos(A+B)-B)

    For example if B=60 degrees and cos(A+B) = 0.5, possible values for cos(A) are 1 and -0.5. In other words values for A of 0 and 240 degrees are both valid.
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  3. #3
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    Re: COS(A+B) Problem - Whats COS A?

    Hi,

    The problem is when you use cos-1 you basically loss info and therefore would like not to use it.

    If you combine


    COS(A+B)=COSACOSB-SINASINB

    And

    COS^2(A)+SIN^2(A)=1

    You get

    Cos(a+b)=cosacosb-(sqrt(1-(cosa)^2))sing

    Which can be simplified with

    A=BC-(sqrt(1-(B)^2))D

    Can anyone here re-arrange this so B=

    Thus no info is lost.

    I would be really impressed if someone could.


    Dave
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  4. #4
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    Re: COS(A+B) Problem - Whats COS A?

    Sorry not

    Cos(a+b)=cosacosb-(sqrt(1-(cosa)^2))sing

    It should read

    Cos(a+b)=cosacosb-(sqrt(1-(cosa)^2))sinb

    Auto correct
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  5. #5
    MHF Contributor ebaines's Avatar
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    Re: COS(A+B) Problem - Whats COS A?

    Using your nomenclature: you have A=BC-(sqrt(1-(B)^2))D, where A, C and D are known values and you want to solve for B. Here A=cos(a+b), B= cos(a), C=cos(b), and D=sin(b):

    A=BC-(\sqrt{1-B^2})D

    Rearrange to get the square root term by itself:

    (\sqrt{1-B^2})D = BC-A

    Square both sdes:

     (1-B^2)D^2 = (BC-A)^2 = B^2C^2- 2BCA + A^2

    Rearrange to get the unknown B into a quadratic equation:

    -B^2(D^2+C^2) +B(2CA) +(D^2-A^2) = 0

    Note that D^2 + C^2 = 1:

    B^2 -B(2CA) + (A^2-D^2) = 0

    Apply the quadratic formula:

     B = \frac { 2CA \pm \sqrt {C^2A^2 -4(A^2-D^2)}}{2}

    Using the example I gave earlier, where cos(a+b) = 0.5 and b = 60 degrees, this solution yields B = cos(a) = 1 or -0.5, exactly the same as before. Doesn't matter whether you use this method or the previous one, you must account for positive and negative values.
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