1. ## Find 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...).

2. 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.

3. 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.

4. Originally Posted by galactus
$\displaystyle \frac{sin(A+B-C)+sin(A-B+C)-sin(A-B-C)+sin^{2}C}{4}$
Sorry but could you explain the first line a little bit.

5. 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.

6. $\displaystyle \sin\alpha\sin\beta\cos\gamma+\sin^2\gamma$
So there is $\displaystyle \cos\gamma$ not $\displaystyle \sin\gamma$!

(And could you please use \sin \cos etc in LaTeX so nobody could even think of variables and it's nicer. Thx)

7. 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.

8. 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.

9. Originally Posted by galactus
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: $\displaystyle sin x$
• "\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).

10. Hey JB. What did you get as a max value?. 9/8?.

$\displaystyle A=B=C=\frac{\pi}{3}$

11. Yes but I'm just guessing - I can't prove it

12. Sub $\displaystyle c={\pi}-A-B$, then you have two variables. Take the derivatives wrt to A and B and whittle it down.

13. $\displaystyle \begin{array}{rcl} \cos{\gamma} &=& \cos{(\pi-(\alpha+\beta))}\\\\ {} &=& -\cos{(\alpha+\beta)}\\\\ {} &=& \sin{\alpha}\sin{\beta}-\cos{\alpha}\cos{\beta}\\\\ \therefore\ \cos^2{\gamma} &=& \sin{\alpha}\sin{\beta}\cos{\gamma}-\cos{\alpha}\cos{\beta}\cos{\gamma} \end{array}$

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 $\displaystyle \frac{9}{8}$.