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Thread: Solving trig equation for its input

  1. #1
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    Solving trig equation for its input

    I have following trig function:
    X = COT-1 ( -(TAN(A) * SIN(B) + SIN(Y) * COS(B)) ) / COS(Y) )
    A and B are known as well as X. I need to solve it for Y.

    I've tried to solve it already, and I have sinking feeling that it might be unsolvable.
    If it helps, A is between -90 and 90 degrees inclusive. B equals to something close to 23.45 degrees, it's the obliquity of earth's ecliptic.

    Thanks.
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  2. #2
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    Re: Solving trig equation for its input

    You can rearrange to get:

    cos(y) cot(x) = -tan(a) sin(b) -sin(y)cos(b)

    Since you have values for x, a, and b, for ease of manipulation we can let cot(x) = P, -tan(a)sin(b) = Q , and -cos(b) = R, and this becomes:

    P cos(y) = Q+Rsin(y)

    Replace the cosine function with its equivalent sine function:

    P sqrt(1-sin^2(y)) = Q + R sin(y).

    Now square both sides:

    P^2 (1- sin^2(y)) = Q^2 + 2QRsin(y) + R^2 sin^2(y)

    Rearrange into a quadratic equation:

    (P^2+R^2) sin^2(y) + 2QR sin(y) +(Q^2-P^2) = 0

    Now use the quadratic formula to solve for sin(y). You'll get two answers - try them both back in the original equation to see if they both work.
    Last edited by ChipB; Mar 22nd 2017 at 08:10 AM.
    Thanks from jacob22
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  3. #3
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    Re: Solving trig equation for its input

    Thanks, it worked wonderfully.
    One thing I would add is that in this case, there seems to be 4 answers that need to be checked. After deriving two answers from quadratic formula and doing ARCSIN, it's important to also check for 180˚- ARCSIN(...) version. The sqrt(1-sin^2(y)) part was quite clever, I missed it.
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