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Math Help - Trigonometry-maximum & minimum values

  1. #1
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    Trigonometry-maximum & minimum values

    For acosA +(-) bsinA +(-) c
    a= rcosA & b= rsinA ( a not = A^o)
    Then a^2+b^2 = r^2-cos^2-A + r^2-sin^2-A = r^2 that= r is sqrt(a^2+b^2)..................................... .......(I could not understand this method. could you explain me about it)
    Last edited by rohith14; October 9th 2009 at 12:44 AM.
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  2. #2
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    I may not understand your question.

    However, to start, r is the hypotenuse of a triangle formed by the axes and an angle.

    a = r cos(A) is true because cosine is the adjacent side (a) over the hypotenuse (r), so this is just solving for a. This is similar with solving for b as well, but it is the side opposite the angle, so we use sine instead of cosine.

    Pythagorean theorem says that for a right triangle a^2 + b^2 = c^2, so when we substitute the above this gives us r^2 cos^2(A) + r^2 sin^2(A).

    If you elaborate more on your question, that would help me.

    Patrick
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  3. #3
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    Trigonometry maximum & minimum vales

    Is there any use of circle with coordinates in the solution and my problem of not understanding is at starting of it after reading the entire proof.
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  4. #4
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    I'm sorry, I don't understand what you are asking.
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  5. #5
    Senior Member pacman's Avatar
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    'Then a^2+b^2 = r^2cos^2-A + r^2sin^2-A = r^2", a pythagoren theorem?

    like this?

    where x = a and y = b?
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  6. #6
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    Trignometry maximum &minimum vales

    "Extreme values of acosA +(-) bcosA +(-) c(anot=A^o) is
    [c - sqrt(a^2+b^2) , c + sqrt(a^2+b^2)]" (I think now it is better) (I edited the qestion could you again read it)
    Last edited by rohith14; October 8th 2009 at 07:05 PM. Reason: grammer problem
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  7. #7
    Senior Member pacman's Avatar
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    same, i can't understand . . . . my problem i think is the same as yours, we both don't learn latexing.
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