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Please suggest suitable approach for this problem

  • 0
    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.2010-11-06
  • 0
    I solved it the answer is 1 and 1/2.2010-11-06
  • 2
    Setting $x=\sin^2\theta$ this reduces to finding the extrema of $x^2+(1-x)^2$ for $x\in[0,1]$.2010-11-06

1 Answers 1

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

  • 0
    Yes that's the observation ;)2010-11-06