DLS
  • DLS
Help finding the maximum value of theta=?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
DLS
  • DLS
Given,if \(\LARGE \theta+\phi=\frac{\pi}{3}\) then find The Maximum Value of\(\Huge (\sin \theta. \sin \phi)\).
DLS
  • DLS
@hartnn
DLS
  • DLS
Attempt: \[\Huge \cos \theta \cos \phi-\sin \theta \sin \phi = \frac{1}{2}\] \[\Huge \sin \theta \sin \phi = \frac{ 2\cos \theta \cos \phi-1}{2}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

DLS
  • DLS
taking cos on both sides ^^
DLS
  • DLS
I guess maximum value of cos theta cos phi is 1?
DLS
  • DLS
In that case, \[\Huge \sin \theta \sin \phi=\frac{1}{2}\]
DLS
  • DLS
only possible when theta=pi/4? hmm..
DLS
  • DLS
but many cases arise..
DLS
  • DLS
@terenzreignz
anonymous
  • anonymous
Simpler: \[\theta=\frac{ \pi }{ 3 }-\phi\]then\[f(\theta, \phi)=\sin(\theta)\sin(\phi)=\sin(\pi/3-\phi)\sin(\phi)\] Find the derivative vs phi:\[\frac{ df(\phi) }{ d \phi }=-\cos(\pi/3-\phi)\sin(\phi)+\sin(\pi/3-\phi)\cos(\phi)=\sin[\pi/3-2\phi]=0\]Thus\[\phi=\pi/6 \rightarrow \theta=\pi/6\]To verify whether it is a mx or min, find the second derivative:\[\frac{ d^2f(\phi) }{ d \phi^2 }=\frac{ dsin(\pi/3-2 \phi) }{ d \phi }=-2\cos(\pi/3-2 \phi) \rightarrow \left[ \frac{ d^2f(\phi) }{ d \phi^2 } \right]_{\phi=\pi/3}=-2<0\]And that means it is a maximum
anonymous
  • anonymous
Mistake, the second derivative is in pi/6 and not pi/3
DLS
  • DLS
yup got it! thanks :D
anonymous
  • anonymous
you are welcome!

Looking for something else?

Not the answer you are looking for? Search for more explanations.