What's the maximum value of $\displaystyle \sin\alpha\sin\beta\cos\gamma+\sin^2\gamma$ if $\displaystyle \alpha+\beta+\gamma=\pi$ (angles of triangle...).

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- Jan 19th 2008, 11:09 AMjames_bondFind the maximum
What's the maximum value of $\displaystyle \sin\alpha\sin\beta\cos\gamma+\sin^2\gamma$ if $\displaystyle \alpha+\beta+\gamma=\pi$ (angles of triangle...).

- Jan 19th 2008, 11:56 AMgalactus
I didn't look at this at length, but you may try expanding it out to:

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

Now, since $\displaystyle A+B+C={\pi}$ perhaps you can simplify it down further. - Jan 19th 2008, 02:09 PMgalactus
I took a closer look at your problem and, if I am correct, it has an interesting

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

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

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

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

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

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

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

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

Unless I made a mistake. Check it out. Interesting solution if correct. - Jan 20th 2008, 02:05 AMjames_bond
- Jan 20th 2008, 05:25 AMgalactus
That is just an expansion/identity of your expression.

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

Since $\displaystyle A+B+C={\pi}$ and $\displaystyle sin({\pi})=0$, I just excluded it. - Jan 20th 2008, 06:49 AMjames_bondQuote:

$\displaystyle \sin\alpha\sin\beta\cos\gamma+\sin^2\gamma$

(And could you please use \sin \cos etc in LaTeX so nobody could even think of variables and it's nicer. Thx) - Jan 20th 2008, 07:09 AMgalactus
DUH. I'm sorry. I misread the problem and it was all for naught. (Fubar)(Headbang)

Let me get back to you.

What do you mean use sin and cos in LaTex?. That's what I've been doing. - Jan 20th 2008, 07:16 AMgalactus
In that event, perhaps we can proceed in the same manner except use:

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

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

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

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

Differentiate wrt A and B and solve. - Jan 20th 2008, 08:23 AMjames_bond
I mean to use \sin command instead of just typing sin.

Here is the difference:

- "sin x" in LaTeX: $\displaystyle sin x$
- "\sin x" in LaTeX: $\displaystyle \sin x$

It's not a big thing but I think that it is nicer (and they wasn't developed the trigonometric commands accidentally). - Jan 20th 2008, 02:19 PMgalactus
Hey JB. What did you get as a max value?. 9/8?.

$\displaystyle A=B=C=\frac{\pi}{3}$ - Jan 21st 2008, 06:01 AMjames_bond
Yes but I'm just guessing - I can't prove it :(

- Jan 21st 2008, 08:55 AMgalactus
Sub $\displaystyle c={\pi}-A-B$, then you have two variables. Take the derivatives wrt to A and B and whittle it down.

- Jan 23rd 2008, 11:08 AMJaneBennet
$\displaystyle \begin{array}{rcl}

\cos{\gamma} &=& \cos{(\pi-(\alpha+\beta))}\\\\

{} &=& -\cos{(\alpha+\beta)}\\\\

{} &=& \sin{\alpha}\sin{\beta}-\cos{\alpha}\cos{\beta}\\\\

\therefore\ \cos^2{\gamma} &=& \sin{\alpha}\sin{\beta}\cos{\gamma}-\cos{\alpha}\cos{\beta}\cos{\gamma}

\end{array}$

So

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

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

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

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