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Find the maximum and minimum values of $\sin^2\theta+\sin^2\phi$ when $\theta+\phi=\alpha$(a constant).


$\theta+\phi=\alpha\implies\phi=\alpha-\theta$
$\sin^2\theta+\sin^2\phi=\sin^2\theta+\sin^2(\alpha-\theta)$

Let $f(\theta)=\sin^2\theta+\sin^2(\alpha-\theta)$
$f'(\theta)=2\sin\theta\cos\theta-2\sin(\alpha-\theta)\cos(\alpha-\theta)$
Putting $f'(\theta)=0$ gives $\sin2\theta=\sin2(\alpha-\theta)$
$2\theta=2\alpha-2\theta\implies \alpha=2\theta$

If $\alpha=2\theta$,then by $\theta+\phi=\alpha$ gives $\phi=\theta$

I am stuck here,the answer given is maximum $1+\cos\alpha$ and minimum $1-\cos\alpha$.But i have found only one critical value(when $\phi=\theta$) and that too i cannot decide whether it will give maximum or minimum value.

Please help.

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  • 1
    $\begingroup$ $\sin 2 \theta = \sin 2 (\alpha -\theta)$ has more than one solution (even modulo $2\pi$). $\endgroup$ – Fabian Jun 1 '16 at 5:02
  • $\begingroup$ and $\cos(2x)=1-2\sin^2(x)$ $\endgroup$ – user175968 Jun 1 '16 at 5:03
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Write \begin{eqnarray} \sin^2 \phi + \sin^2 \theta &=& \sin^2 \phi + 1-\cos^2 \theta \\ &=& 1+(\sin \phi - \cos \theta) (\sin \phi + \cos \theta) \\ &=& 1+(\sin \phi - \sin ( { \pi \over 2} - \theta)) (\sin \phi + \sin ( { \pi \over 2} - \theta)) \\ &=& 1+4 \cos ({ \phi -\theta + { \pi \over 2}\over 2} ) \sin ({ \phi +\theta - { \pi \over 2}\over 2} ) \sin ({ \phi -\theta + { \pi \over 2}\over 2} ) \cos ({ \phi +\theta - { \pi \over 2}\over 2} ) \\ &=& 1+ \sin (\phi -\theta + { \pi \over 2}) \sin ( \phi +\theta - { \pi \over 2} ) \\ &=& 1 - \sin (\phi -\theta + { \pi \over 2}) \cos \alpha \end{eqnarray} We can choose $\phi -\theta$ to have arbitrary values while maintaining the relationship $\phi+\theta = \alpha$, and so the above can be written as $1-\sin t \cos \alpha$, which can be easily extremized,

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