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Please suggest suitable approach for this problem

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In which interval? It's periodic with period $\frac{\pi}{2}$, so it will have a lot of maxima and minima. But if you're restricting to $\left[0,\frac{\pi}{2}\right]$, then it has one minimum and two maxima at the ends. –  J. M. Nov 6 '10 at 9:59
    
I solved it the answer is 1 and 1/2. –  Trewick Marian Nov 6 '10 at 10:10
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Setting $x=\sin^2\theta$ this reduces to finding the extrema of $x^2+(1-x)^2$ for $x\in[0,1]$. –  Robin Chapman Nov 6 '10 at 10:26
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up vote 6 down vote accepted

Note that $$1 = (\sin^2\theta + \cos^2\theta)^2$$ and use $\sin2\theta = 2\sin\theta\cos\theta.$

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Yes that's the observation ;) –  Trewick Marian Nov 6 '10 at 10:10
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