Question

Help finding the maximum value of theta=?

Answer 1

Given,if \(\LARGE \theta+\phi=\frac{\pi}{3}\) then find The Maximum Value of\(\Huge (\sin \theta. \sin \phi)\).

Answer 2

@hartnn

Answer 3

Attempt:
\[\Huge \cos \theta \cos \phi-\sin \theta \sin \phi = \frac{1}{2}\]
\[\Huge \sin \theta \sin \phi = \frac{ 2\cos \theta \cos \phi-1}{2}\]

Answer 4

taking cos on both sides ^^

Answer 5

I guess maximum value of cos theta cos phi is 1?

Answer 6

In that case,
\[\Huge \sin \theta \sin \phi=\frac{1}{2}\]

Answer 7

only possible when theta=pi/4?
hmm..

Answer 8

but many cases arise..

Answer 9

@terenzreignz

Answer 10

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

Answer 11

Mistake, the second derivative is in pi/6 and not pi/3

Answer 12

yup got it! thanks :D

Answer 13

you are welcome!