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Extreme values and Periodicity of Trigonometric functions

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Question

The maximum value of $1 + \sin (\frac{π}{4} + θ) + 2 \cos (\frac{π}{4} - θ)$ for real values of $θ$ , is

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Solution

We have,
$\begin{aligned} \Rightarrow & y & = 1 + \sin (\frac{π}{4} + θ) + 2 \cos (\frac{π}{4} - θ) \\ \Rightarrow & y & = 1 + \frac{1}{\sqrt{2}} (\sin θ + \cos θ) + \sqrt{2} (\cos θ + \sin θ) \\ \Rightarrow & y & = 1 + (\sqrt{2} + \frac{1}{\sqrt{2}}) sinθ + (\sqrt{2} + \frac{1}{\sqrt{2}}) cosθ \\ \Rightarrow & y & = 1 + (\sqrt{2} + \frac{1}{\sqrt{2}}) (cosθ + sinθ) (∵ \cos θ + \sin θ \leq \sqrt{2}] \\ \Rightarrow & y \leq 1 + (\sqrt{2} + \frac{1}{\sqrt{2}}) & \sqrt{2} \Rightarrow y \leq 4 \end{aligned}$
Hence, the maximum value is 4.

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