Question

If the maximum value of $3cos\theta +5sin\left(\theta -\frac{\pi }{6}\right)$ for any real value of $\theta$ is $\sqrt{k}$, than $k$ =

Answer 1

Answer: 19

Let, the functions is,
$f\left(\theta \right)=3cos\theta +5sin\theta \cdot cos\frac{\pi }{6}-5sin\frac{\pi }{6}cos\theta$
$=3cos\theta +5\times \frac{\sqrt{3}}{2}sin\theta -5\times \frac{1}{2}cos\theta$
$=\left(3-\frac{5}{2}\right)cos\theta +5\times \frac{\sqrt{3}}{2}sin\theta$
$=\frac{1}{2}cos\theta +\frac{5\sqrt{3}}{2}sin\theta$
$=\frac{1}{2}cos\theta +\frac{5\sqrt{3}}{2}sin\theta$
max $f\left(\theta \right)=\sqrt{\frac{1}{4}+\frac{25}{4}\times 3}=\sqrt{\frac{76}{4}}=\sqrt{19}$