#Trigo-2

The question demands for the Maximum and minimum value of the above function.

Lemma: If there is a function of the form $a \cos \theta + b \sin \theta + c$ , then it falls in the range of

$\sqrt{a^2+b^2}+c \geq a \cos \theta + b \sin \theta + c \geq -\sqrt{a^2+b^2}+c$

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Solution

We have $5 \cos \theta + 3 \cos (\dfrac{\pi}{3}+\theta)+3$

$\implies 5 \cos \theta + 3 \cos \theta \cos \dfrac{\pi}{3} - 3 \sin \theta \sin \dfrac{\pi}{3} + 3$

$\implies 5 \cos \theta + 3 \cos \theta \times \dfrac{1}{2} - 3 \sin \theta \times \dfrac{\sqrt{3}}{2} + 3$

$\implies \dfrac{13}{2} \cos \theta - \dfrac{3\sqrt{3}}{2} \sin \theta + 3$ [ $\color{#D61F06} \text{Brought down to the form a cos x + b sin x + c}$ ]

$\implies \sqrt{ \dfrac{13^2}{2^2} + \dfrac{27}{2^2}} +3 \geq 5 \cos \theta + 3 \cos (\dfrac{\pi}{3}+\theta)+3 \geq -\sqrt{ \dfrac{13^2}{2^2} + \dfrac{27}{2^2}} +3$

$\implies 7 + 3 \geq 5 \cos \theta + 3 \cos (\dfrac{\pi}{3}+\theta)+3 \geq -7+3$

$\implies 10 \geq 5 \cos \theta + 3 \cos (\dfrac{\pi}{3}+\theta)+3 \geq -4$

So Range = $[-4, 10]$

And hence answer is $\color{#D61F06}{[-4,10]}$