Question

Find $a$ and $b$ such that for all $x,a\le 3\cos x+5\sin (x-\frac{\pi }{6})\le b$.

Answer 1

$\Rightarrow 3\cos x+5\sin (x-\frac{\pi }{6})$

$=3\cos x+5\sin x\cos \frac{\pi }{6}-5\cos x\sin \frac{\pi }{6}$

$=(3-\frac{5}{2})\cos x+\frac{5\sqrt{3}}{2}\sin x$.

$=\frac{\cos x}{2}+\frac{5\sqrt{3}}{2}\sin x$.

The maximum & minimum value of $a\cos x+b\sin x=\pm \sqrt{a^2+b^2}$

$=\pm \sqrt{{\left(\frac{1}{2}\right)}^2+{\left(\frac{5\sqrt{3}}{2}\right)}^2}=\pm \sqrt{\frac{1}{4}+\frac{75}{4}}\Rightarrow \pm \sqrt{19}$.

$\therefore -\sqrt{19}\le 3\cos x+5\sin (x-\frac{\pi }{6})\le \sqrt{19}$

$\therefore a=-\sqrt{19},b=\sqrt{19}$.