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...).
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.
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.
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.
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).
$\displaystyle \begin{array}{rcl}
\cos{\gamma} &=& \cos{(\pi-(\alpha+\beta))}\\\\
{} &=& -\cos{(\alpha+\beta)}\\\\
{} &=& \sin{\alpha}\sin{\beta}-\cos{\alpha}\cos{\beta}\\\\
\therefore\ \cos^2{\gamma} &=& \sin{\alpha}\sin{\beta}\cos{\gamma}-\cos{\alpha}\cos{\beta}\cos{\gamma}
\end{array}$
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 $\displaystyle \frac{9}{8}$.