Question

Find the maximum value of cos$\theta$ + 3$\sqrt{2}$ sin $(\theta +\frac{\pi }{4})$

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Answer 1

We will first simplify the expression and write it in the form a sin$\theta$ + bcos$\theta$

We know its maximum value is $\sqrt{a^2+b^2}$

cos$\theta$ + 3$\sqrt{2}$ sin $(\theta +\frac{\pi }{4})$ = cos$\theta$ + 3$\sqrt{2}$ $(sin\theta \times \frac{1}{\sqrt{2}}+\frac{cos\theta }{\sqrt{2}})$

= $cos\theta$ + 3 $sin\theta$ + 3$cos\theta$

= 4 $cos\theta$ + 3 $sin\theta$

Maximum value of its expression is $\sqrt{3^2+4^2}$

= 5