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Math Help - 2 Trigonometric Equations 2 Unkowns

  1. #1
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    2 Trigonometric Equations 2 Unkowns

    I can't remember how to solve these. Any help? I tried using Euler's formula, no avail.

    sin(x)+sin(y) = 0.90154195
    cos(x)+cos(y) = 1.64483049

    Solve for x & y:
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  2. #2
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    Quote Originally Posted by RainmanTrail View Post
    I can't remember how to solve these. Any help? I tried using Euler's formula, no avail.

    sin(x)+sin(y) = 0.90154195
    cos(x)+cos(y) = 1.64483049

    Solve for x & y:
    Actually you can use Euler's
    (1) : sin(x)+sin(y) = 0.90154195
    (2) : cos(x)+cos(y) = 1.64483049

    (2) + i(1) : e^ix + e^iy = 1.64 + i0.90
    (2) - i(1) : e^-ix + e^-iy = 1.64 - i0.90

    Now use A = e^ix; B = e^iy.

    A + B = 1.64 + i0.90
    1/A + 1/B = 1.64 - i0.90

    Solve this system.
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  3. #3
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    Quote Originally Posted by snowtea View Post
    Actually you can use Euler's
    (1) : sin(x)+sin(y) = 0.90154195
    (2) : cos(x)+cos(y) = 1.64483049

    (2) + i(1) : e^ix + e^iy = 1.64 + i0.90
    (2) - i(1) : e^-ix + e^-iy = 1.64 - i0.90

    Now use A = e^ix; B = e^iy.

    A + B = 1.64 + i0.90
    1/A + 1/B = 1.64 - i0.90

    Solve this system.
    Thanks, I set up the equation incorrectly, I was getting 0=ln(3.28). It's been a while since I've worked with this stuff, from here would I just substitute for A or B and solve the equations? The i is throwing me off a bit.
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  4. #4
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    Quote Originally Posted by RainmanTrail View Post
    Thanks, I set up the equation incorrectly, I was getting 0=ln(3.28). It's been a while since I've worked with this stuff, from here would I just substitute for A or B and solve the equations? The i is throwing me off a bit.
    Yes, you can substitute, and solve it like any other equation.

    If the i is tripping you up, there is another way to solve it directly:
    Square the first and second equations and add them (it simplifies very nicely with trig identities).
    You will be able to solve for x - y.
    Then use substitution.
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  5. #5
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    I'm attempting to solve by squaring the equations and adding them. I come up with this:
    sin^2(x)+2sin(x)sin(y)+sin^2(y)+cos^2(x)+2cos(x)co s(y)+cos^2(y) = (0.90)^2+(1.64)^2
    since sin^2(x)+cos^2(x)=1, we have:
    1+1+2sin(x)sin(y)+2cos(x)cos(y)=...
    which simplifies as:
    2+2cos(x-y) = 0.9^2 + 1.64^2

    is this where you were headed? I'm not sure how to solve for x and y here though...
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  6. #6
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    Just solve to get x - y = C (some constant or family of constants).
    Then substitue x = y + C into one of the 2 original equations.
    Use trig rules to decompose sin(y+C) or cos(y+C).
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