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Math Help - Trigonomic function question

  1. #1
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    Trigonomic function question

    Hey, I have been working a bit with the trig functions and came across this problem:

    B = (A + sin(theta) )/ (sin(theta)*cos(theta))

    when you separate the above fraction into the following:

    C = A/(sin(theta)*cos(theta))

    D = sin(theta) / (sin(theta)*cos(theta)) => 1/ cos(theta)

    C+D != B if theta = 0 as sin(0) = 0 and cos(0) = 1

    I came across this problem when trying to solve :

    a - b*cos(theta) - b^2*sin(theta) = 0

    Is there any method to make sure that this doesnt happen or do you just have to check the formula through for this?

    Thanks for any replies.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by FireSoul View Post
    Hey, I have been working a bit with the trig functions and came across this problem:

    B = (A + sin(theta) )/ (sin(theta)*cos(theta))

    when you separate the above fraction into the following:

    C = A/(sin(theta)*cos(theta))

    D = sin(theta) / (sin(theta)*cos(theta)) => 1/ cos(theta)

    C+D != B if theta = 0 as sin(0) = 0 and cos(0) = 1

    I came across this problem when trying to solve :

    a - b*cos(theta) - b^2*sin(theta) = 0

    Is there any method to make sure that this doesnt happen or do you just have to check the formula through for this?

    Thanks for any replies.
    I would simply note that none of B, C, or D exist when \theta = 0, so there really is no problem.

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by FireSoul View Post
    I came across this problem when trying to solve :

    a - b*cos(theta) - b^2*sin(theta) = 0
    You are trying to solve for \theta? My recommendation:
    b^2sin(\theta) = a - b~cos(\theta)<-- Square both sides

    b^4sin^2(\theta) = a^2 - 2ab~cos(\theta) + b^2 cos^2(\theta)

    Now, sin^2(\theta) = 1 - cos^2(\theta), so
    b^4(1 - cos^2(\theta)) = a^2 - 2ab~cos(\theta) + b^2 cos^2(\theta)

    b^4 - b^4cos^2(\theta) = a^2 - 2ab~cos(\theta) + b^2 cos^2(\theta)

    (b^4 + b^2)cos^2(\theta) - 2ab~cos(\theta) + (a^2 - b^4) = 0

    Now let y = cos(\theta). Then your equation is
    (b^4 + b^2)y^2 - 2aby + (a^2 - b^4) = 0
    which is a quadratic you can solve for y.

    Note: When doing this you MUST check your solutions for \theta in the original equation for \theta as the method I am using here is likely to add spurious solutions.

    -Dan
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