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Math Help - please help me find the unknowns!

  1. #1
    maa
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    3.5 cos(x) + 2 cos (y) = 4
    3.5 sin (x) = 2 sin (y) + 0.5

    I already have the final answer which are x=36.87 and y=53.13
    I need to know how they solved it.

    thanks!!

    oh and i've tried everything. adding them,dividing them,multiplying with cos(x),sin(x)..etc everything!;/
    Last edited by mr fantastic; October 31st 2009 at 04:04 PM. Reason: Merged posts
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  2. #2
    Flow Master
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    Quote Originally Posted by maa View Post
    3.5 cos(x) + 2 cos (y) = 4
    3.5 sin (x) = 2 sin (y) + 0.5

    I already have the final answer which are x=36.87 and y=53.13
    I need to know how they solved it.

    thanks!!

    oh and i've tried everything. adding them,dividing them,multiplying with cos(x),sin(x)..etc everything!;/
    Your problem essentially boils down to solving the following four equations simultaneously:

    3.5a + 2b = 4 => 7a + 4b = 8 .... (1)

    3.5c - 2d = 0.5 => 7c - 4d = 1 .... (2)

    a^2 + c^2 = 1 .... (3)

    b^2 + d^2 = 1 .... (4)

    where all of the a, b, c and d lie between -1 and 1 inclusive.
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  3. #3
    maa
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    ok I understand how you got the 4 equations but how can i solve it? using a matrix?

    7 4 0 0 8
    0 0 7 -4 1
    ? 0 ? 0 1
    0 ? 0 ? 1

    is that right? if yes, what should i put in ( ? ) cause the a,b,c,d are square in this case. (is it square root to 1 which equals one?)
    thanks again!
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  4. #4
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    Quote Originally Posted by maa View Post
    ok I understand how you got the 4 equations but how can i solve it? using a matrix?

    7 4 0 0 8
    0 0 7 -4 1
    ? 0 ? 0 1
    0 ? 0 ? 1

    is that right? if yes, what should i put in ( ? ) cause the a,b,c,d are square in this case. (is it square root to 1 which equals one?)
    thanks again!
    Marices can only be used to solve a system of linear equations. These equations aren't all linear. I'd substitute (1) and (2) into (3) and (4) to get two quadratic equations in a and d. Get them and solve them.
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