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Extreme values and Periodicity of Trigonometric functions

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Question

The maximum value of the expression $\frac{1}{\sin^{2} θ + 3 \sin θcos θ + 5 \cos^{2} θ} is$

Moderate

Solution

$\begin{array}{l} Let f (θ) = \frac{1}{\sin^{2} θ + 3 \sin θcos θ + 5 \cos^{2} θ} \\ \begin{aligned} Again let g (θ) & = \sin^{2} θ + 3 \sin θcos θ + 5 \cos^{2} θ \\ = \frac{1 - \cos 2 θ}{2} + 5 (\frac{1 + \cos 2 θ}{2}) + \frac{3}{2} \sin 2 θ \\ = 3 + 2 \cos 2 θ + \frac{3}{2} \sin 2 θ \end{aligned} \\ ∴ g (θ)_{min} = - \sqrt{4 + \frac{9}{4}} = 3 - \frac{5}{2} = \frac{1}{2} \\ ∴ f (θ) = \frac{1}{g (θ)_{min}} = 2 \end{array}$

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