MHB Finding Max and Min of $sin(x)sin(y)$ given $x+y=\dfrac {2\pi}{3}$

[Albert1]

$x+y=\dfrac {2\pi}{3}$

$0\leq x\leq \dfrac{\pi}{2}$

$find: $

$(1):max(sin(x)sin(y))$

$(2):min(sin(x)sin(y))$

 

[kaliprasad]

$x+y=\dfrac {2\pi}{3}$

$0\leq x\leq \dfrac{\pi}{2}$

$find: $

$(1):max(sin(x)sin(y))$

$(2):min(sin(x)sin(y))$

Spoiler

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 $x= 0, y = \frac{2\pi}{3}$

 

[Albert1]

Spoiler

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 $x= 0, y = \frac{2\pi}{3}$

Spoiler

(*)=
$\sin\,x \sin\, y = \sin ^2 \frac {\pi}{3} - \sin ^2 \alpha = \frac{3}{4} - \sin ^2 \alpha$-----(*)

 

[kaliprasad]

Spoiler

(*)=
$\sin\,x \sin\, y = \sin ^2 \frac {\pi}{3} - \sin ^2 \alpha = \frac{3}{4} - \sin ^2 \alpha$-----(*)

Thanks done the needful