Albert said:$x+y=\dfrac {2\pi}{3}$
$0\leq x\leq \dfrac{\pi}{2}$
$find: $
$(1):max(sin(x)sin(y))$
$(2):min(sin(x)sin(y))$
kaliprasad said:we have
$\sin (A+B) \sin (A-B) = (\sin\, A \cos\, B + \cos \, A \sin \, B) (\sin\, A \cos\, B - \cos \, A \sin \, B)$
$=(\sin^2 A \cos^2 B - \cos^2 A \sin^2 B) = \sin ^2 A (1- sin ^2 B ) - \sin ^2 B ( 1 - sin ^2 A) = \sin ^2 A - \sin ^2 B$
let $x= \frac {\pi}{3} + \alpha$ and hence $y = \frac {\pi}{3} - \alpha$
so
$\sin\, \sin\, y = \sin ^2 \frac {\pi}{3} - \sin ^2 \alpha = \frac{3}{4} \sin ^2 \alpha$-----(*)
the above is maximum when $\sin^2 \alpha = 0$ minimum or $x = y = \frac {\pi}{3}$ and value is $\frac{3}{4}$
now both A and B are 0 or positive and minumum is zero when $x= 0, y = \frac{2\pi}{3}$
Albert said:(*)=
$\sin\,x \sin\, y = \sin ^2 \frac {\pi}{3} - \sin ^2 \alpha = \frac{3}{4} - \sin ^2 \alpha$-----(*)
The maximum value of $sin(x)sin(y)$ given $x+y=\dfrac {2\pi}{3}$ is $\dfrac {3\sqrt{3}}{4}$.
The minimum value of $sin(x)sin(y)$ given $x+y=\dfrac {2\pi}{3}$ is $-\dfrac {3\sqrt{3}}{4}$.
To find the maximum and minimum values of $sin(x)sin(y)$ given $x+y=\dfrac {2\pi}{3}$, we can use the following steps:
No, there can only be one maximum and one minimum value of $sin(x)sin(y)$ given $x+y=\dfrac {2\pi}{3}$. This is because the function is continuous and the domain is limited to a specific range, so there can only be one extreme value.
The value of $x+y$ does not affect the maximum and minimum values of $sin(x)sin(y)$ itself, but it does determine the range of possible values. In this case, since $x+y=\dfrac {2\pi}{3}$, the maximum and minimum values will fall within the range of $-\dfrac {3\sqrt{3}}{4}$ to $\dfrac {3\sqrt{3}}{4}$.