# Find the maximum

• January 19th 2008, 11:09 AM
james_bond
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...).
• January 19th 2008, 11:56 AM
galactus
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.
• January 19th 2008, 02:09 PM
galactus
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.
• January 20th 2008, 02:05 AM
james_bond
Quote:

Originally Posted by galactus
$\frac{sin(A+B-C)+sin(A-B+C)-sin(A-B-C)+sin^{2}C}{4}$

Sorry but could you explain the first line a little bit.:o
• January 20th 2008, 05:25 AM
galactus
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.
• January 20th 2008, 06:49 AM
james_bond
Quote:

$\sin\alpha\sin\beta\cos\gamma+\sin^2\gamma$
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)
• January 20th 2008, 07:09 AM
galactus
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.
• January 20th 2008, 07:16 AM
galactus
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.
• January 20th 2008, 08:23 AM
james_bond
Quote:

Originally Posted by galactus
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.

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).
• January 20th 2008, 02:19 PM
galactus
Hey JB. What did you get as a max value?. 9/8?.

$A=B=C=\frac{\pi}{3}$
• January 21st 2008, 06:01 AM
james_bond
Yes but I'm just guessing - I can't prove it :(
• January 21st 2008, 08:55 AM
galactus
Sub $c={\pi}-A-B$, then you have two variables. Take the derivatives wrt to A and B and whittle it down.
• January 23rd 2008, 11:08 AM
JaneBennet
$\begin{array}{rcl}
\cos{\gamma} &=& \cos{(\pi-(\alpha+\beta))}\\\\
{} &=& -\cos{(\alpha+\beta)}\\\\
{} &=& \sin{\alpha}\sin{\beta}-\cos{\alpha}\cos{\beta}\\\\
\therefore\ \cos^2{\gamma} &=& \sin{\alpha}\sin{\beta}\cos{\gamma}-\cos{\alpha}\cos{\beta}\cos{\gamma}
\end{array}$

So

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

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

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

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 $\frac{9}{8}$.