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Math Help - Find the maximum

  1. #1
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    Find the maximum

    What's the maximum value of \sin\alpha\sin\beta\cos\gamma+\sin^2\gamma if \alpha+\beta+\gamma=\pi (angles of triangle...).
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  2. #2
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    I didn't look at this at length, but you may try expanding it out to:

    \frac{sin(A+B-C)+sin(A-B+C)-sin(A-B-C)+sin^{2}C}{4}

    Now, since A+B+C={\pi} perhaps you can simplify it down further.
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  3. #3
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    I took a closer look at your problem and, if I am correct, it has an interesting

    max and min. I got \frac{\sqrt{5}+1}{8} as the max and

    \frac{1-\sqrt{5}}{8} as the min. It's related to The Golden Ratio.


    \frac{sin(A-B+C)+sin(-A+B+C)+sin(A+B-C)+sin^{2}C}{4}

    =\frac{sin(A-({\pi}-A))+sin(-A+({\pi}-A))+sin({\pi}-C-C)+sin^{2}C}{4}

    =\frac{sin(2A-{\pi})+sin({\pi}-2A)+sin({\pi}-2C)+sin^{2}C}{4}

    =\frac{sin(2C)+sin^{2}C}{4}

    We can see from differentiating or graphing that the max occurs at

    \frac{\sqrt{5}+1}{8}

    Unless I made a mistake. Check it out. Interesting solution if correct.
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  4. #4
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    Quote Originally Posted by galactus View Post
    \frac{sin(A+B-C)+sin(A-B+C)-sin(A-B-C)+sin^{2}C}{4}
    Sorry but could you explain the first line a little bit.
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  5. #5
    Eater of Worlds
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    That is just an expansion/identity of your expression.

    sinAsinBsinC+sin^{2}C =\frac{sin(A-B+C)-sin(A-B-C)+sin(A+B-C)-sin(A+B+C)+sin^{2}C}{4}

    Since A+B+C={\pi} and sin({\pi})=0, I just excluded it.
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  6. #6
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    \sin\alpha\sin\beta\cos\gamma+\sin^2\gamma
    So there is \cos\gamma not \sin\gamma!

    (And could you please use \sin \cos etc in LaTeX so nobody could even think of variables and it's nicer. Thx)
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  7. #7
    Eater of Worlds
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    DUH. I'm sorry. I misread the problem and it was all for naught.

    Let me get back to you.

    What do you mean use sin and cos in LaTex?. That's what I've been doing.
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  8. #8
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    In that event, perhaps we can proceed in the same manner except use:

    sin(A)sin(B)cos(C)+sin^{2}(C)

    =\frac{cos(A-B-C)+cos(A-B+C)-cos(A+B-C)+1-2(cos(2C)-1)}{4}

    or let C={\pi}-A-B and use:

    F(A,B)=sin(A)sin(B)cos({\pi}-A-B)+sin^{2}({\pi}-A-B)

    Differentiate wrt A and B and solve.
    Last edited by galactus; January 20th 2008 at 03:23 PM.
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  9. #9
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    Quote Originally Posted by galactus View Post
    DUH. I'm sorry. I misread the problem and it was all for naught.

    Let me get back to you.

    What do you mean use sin and cos in LaTex?. That's what I've been doing.
    I mean to use \sin command instead of just typing sin.
    Here is the difference:
    • "sin x" in LaTeX: sin x
    • "\sin x" in LaTeX: \sin x

    It's not a big thing but I think that it is nicer (and they wasn't developed the trigonometric commands accidentally).
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  10. #10
    Eater of Worlds
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    Hey JB. What did you get as a max value?. 9/8?.

    A=B=C=\frac{\pi}{3}
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  11. #11
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    Yes but I'm just guessing - I can't prove it
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  12. #12
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    Sub c={\pi}-A-B, then you have two variables. Take the derivatives wrt to A and B and whittle it down.
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  13. #13
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    \begin{array}{rcl}<br />
\cos{\gamma} &=& \cos{(\pi-(\alpha+\beta))}\\\\<br />
{} &=& -\cos{(\alpha+\beta)}\\\\<br />
{} &=& \sin{\alpha}\sin{\beta}-\cos{\alpha}\cos{\beta}\\\\<br />
\therefore\ \cos^2{\gamma} &=& \sin{\alpha}\sin{\beta}\cos{\gamma}-\cos{\alpha}\cos{\beta}\cos{\gamma}<br />
\end{array}

    So

    \color{white}.\ \ . \sin{\alpha}\sin{\beta}\cos{\gamma}+\sin^2{\gamma}

    =\ \cos^2{\gamma}+\cos{\alpha}\cos{\beta}\cos{\gamma}  +\sin^2{\gamma}

    =\ 1+\cos{\alpha}\cos{\beta}\cos{\gamma}

    The maximum value of \cos{\alpha}\cos{\beta}\cos{\gamma} is \frac{1}{8}. Hence the maximum value of 1+\cos{\alpha}\cos{\beta}\cos{\gamma} is \frac{9}{8}.
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