Question

Find the stationary values of sinx sin y sin (x+y).

To get stationary values, we solve f_x=0 and f_y=0 simultaneously.
f_x=0= cos x sin y sin (x+y)+sin x sin y cos (x+y)
f_y=0= sin x cos y sin (x+y)+sin x sin y cos (x+y)
Adding the above 2 equations,
sin (x+y) [cos x sin y+sin x cos y ]+cos (x+y) sin x sin y=0
sin^2 (x+y)+cos (x+y) sin x sin y=0
….
what next ?
or any other better approach ?

Answer 1

@SithsAndGiggles

Answer 2

\[\cos(x)\sin(y)\sin(x+y)=\sin(x)\cos(y)\sin(x+y) \\ \sin(x+y)(\sin(x)\cos(y)-\sin(y)\cos(x))=0 \\ \sin(x+y)\sin(x-y)=0\]

Answer 3

set both factors =0

Answer 4

sin(x+y)=0 or sin(x-y)=0

Answer 5

@hartnn any questions? I think you taught to hard.

Answer 6

too*

Answer 7

i got that, but there will be other values too , right ?
i need all the stationary values.

from that i get
x+y =0, x-y =0
hence x=y=0

Answer 8

I get x+y=npi and x-y=npi
therefore x=npi and y=0

Answer 9

\[2x+0=2n \pi => x=n \pi => y=0 \]

Answer 10

http://www.wolframalpha.com/input/?i=stationary+values+of+sin+x++sin+y++sin+%28x%2By%29

Answer 11

sin(x+y)=0
sin(u)=0 when u=npi so x+y=npi
sin(x-y)=0
sin(v)=0 when v=npi so x-y=npi
_______________________add equations
2x=2npi

Answer 12

-2pi/3 , 2pi/3
-2pi/3 , pi/3
and all other values....how would i get them ?

Answer 13

hmm...

Answer 14

-x-y=npi
x-y=npi
---------
-2y=2npi
and I guess we can also get y=-npi when x=0
but for the other values thinking

Answer 15

\[0=\sin(x+y)(\sin(x)\cos(y)+\sin(y)\cos(x)) \\+\cos(x+y)(\sin(x)\sin(y)+\sin(x)\sin(y)) \\ 0=\sin(x+y)\sin(x+y)+\cos(x+y)2\sin(x)\sin(y)\]

Answer 16

i notice in your equation above you don't have that 2 constant multiple for the second addend

Answer 17

oh yeah, missed it!

Answer 18

sin^2 (x+y)+2 cos (x+y) sin x sin y=0

Answer 19

\[0=1-\cos^2(x+y)+\cos(x+y)2\sin(x)\sin(y) \\ \cos^2(x+y)-\cos(x+y)2\sin(x)\sin(y)-1=0\]
it would be easier if that 2sin(x)sin(y) wasn't there :(

Answer 20

yeah, a quadratic in cos (x+y)
but that would still have given values of x+y

Answer 21

\[\sin(x)\sin(y)=\frac{1-\cos^2(x+y)}{-2\cos(x+y)}=\frac{-\sin^2(x+y)}{2\cos(x+y)}\]
\[\frac{-\sin^2(x+y)}{2\cos(x+y)} \sin(x+y)=0\]
This doesn't work either ...
I tried solving our equation for sin(x)sin(y) and pluggin it into the original
where we sit the original=0
but this still only gives us sin(x+y)=0

Answer 22

\[\cos(x+y)=\frac{-2\sin(x)\sin(y) \pm \sqrt{4\sin^2(x)\sin^2(y)+4}}{2}\]
so
\[\cos(x+y)=\frac{-2\sin(x)\sin(y) \pm 2 \sqrt{\sin^2(x)\sin^2(y)+1}}{2}\]
\[\cos(x+y)=-\sin(x)\sin(y) \pm \sqrt{(\sin(x)\sin(y))^2+1}\]

I don't think I know what to do with this if i can do anything with it
just thinking

Answer 23

I wasn't sure if anyone brought it up, but we could use an identity for the \(f_x=0\) equation:
\[\begin{align*}
f_x&=\cos x\sin y\sin (x+y)+\sin x\sin y\cos(x+y)\\
0&=\cos x\sin (x+y)+\sin x \cos (x+y)\\
0&=\sin(2x+y)&\text{(angle sum id.)}
\end{align*}\]
Similarly for \(f_y=0\),
\[\begin{align*}
f_y&=\sin x\cos y\sin (x+y)+\sin x\sin y\cos(x+y)\\
0&=\cos y\sin (x+y)+\sin y\cos(x+y)\\
0&=\sin(x+2y)&\text{(angle sum id.)}
\end{align*}\]
So we have
\[\begin{cases}
2x+y=n\pi\\
x+2y=n\pi
\end{cases}~~\implies~~x=y\]
hmm, does that help with anything?

Answer 24

yeah he already has the x=y
he wants to other solutions wolfram gave

Answer 25

but i like the way you found that solution it is kinda cuter than what i did

Answer 26

well I think 2x+y=npi
and x+2y=npi
actually gives us your other solutions @hartnn
2x+y=npi
-2x-4y=-2npi
-------------------------
-3y=-2npi
y=2npi/3 then x=npi-2(2npi/3)=-npi/3

Answer 27

awesome!
thank you very much :)