# If x+y=2pi/4, find maximum and minimum of sinx+siny?

## When $x + y = \frac{2 \pi}{4} , x \ge 0 , y \ge 0 :$ the maximum and the minimum value of  sinx+siny=?

Maximum $= \setminus \sqrt{2}$

Minimum $= - \setminus \sqrt{2}$

#### Explanation:

Given that

$x + y = \frac{2 \setminus \pi}{4}$

$x + y = \setminus \frac{\pi}{2}$

$y = \setminus \frac{\pi}{2} - x$

Now, setting $y = \setminus \frac{\pi}{2} - x$ in given function as follows

$f \left(x , y\right) = \setminus \sin x + \setminus \sin y$

$f \left(x\right) = \setminus \sin x + \setminus \sin \left(\setminus \frac{\pi}{2} - x\right)$

$f \left(x\right) = \setminus \sin x + \setminus \cos x$

$f ' \left(x\right) = \setminus \cos x - \setminus \sin x$

$f ' ' \left(x\right) = - \sin x - \cos x$

For minima & maxima, $f ' \left(x\right) = 0$

$\setminus \therefore \setminus \cos x - \setminus \sin x = 0$

$\setminus \tan x = 1$

$x = n \setminus \pi + \setminus \frac{\pi}{4}$

$\setminus \implies f ' ' \left(\setminus \frac{\pi}{4}\right) = - \setminus \sin \left(\setminus \frac{\pi}{4}\right) - \setminus \cos \left(\setminus \frac{\pi}{4}\right) = - \setminus \sqrt{2} < 0$

hence, function is maximum at $x = \setminus \frac{\pi}{4}$ & the maximum value is given as

$f \left(\setminus \frac{\pi}{4}\right) = \setminus \sin \left(\setminus \frac{\pi}{4}\right) + \setminus \cos \left(\setminus \frac{\pi}{4}\right) = \setminus \sqrt{2}$

$\setminus \implies f ' ' \left(\frac{5 \setminus \pi}{4}\right) = - \setminus \sin \left(\frac{5 \setminus \pi}{4}\right) - \setminus \cos \left(\frac{5 \setminus \pi}{4}\right) = \setminus \sqrt{2} > 0$

hence, function is minimum at $x = \frac{5 \setminus \pi}{4}$ & the minimum value is given as

$f \left(\frac{5 \setminus \pi}{4}\right) = \setminus \sin \left(\frac{5 \setminus \pi}{4}\right) + \setminus \cos \left(\frac{5 \setminus \pi}{4}\right) = - \setminus \sqrt{2}$

Jul 27, 2018

$x + y = \frac{2 \pi}{4} \implies y = \frac{\pi}{2} - x$

$\therefore \sin x + \sin y = \sin x + \sin \left(\frac{\pi}{2} - x\right)$

$= \sin x + \cos x$

$= \sqrt{2} \left(\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x\right)$

$= \sqrt{2} \sin \left(x + \frac{\pi}{4}\right)$

So, maximum and minimum of "$\sin x + \sin y$":

• $\sqrt{2}$

• $- \sqrt{2}$