Question

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

Answer 1

Let, the functions is,
$f \left(\theta\right)=3\,cos\,\theta+5\,sin\,\theta\cdot cos \frac{\pi}{6}-5\,sin \frac{\pi}{6} cos\,\theta $
$=3\, cos\,\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 $f \left(\theta\right)=\sqrt{\frac{1}{4}+\frac{25}{4}\times3}=\sqrt{\frac{76}{4}}=\sqrt{19}$