Question

Find minimum and maximum value of $3\cos x+5\sin (x-\pi /6)$

• A. $-\sqrt{25},\sqrt{25}$
• B. -17,17
• C. $-\sqrt{21},\sqrt{21}$
• D. None of these

Answer 1

Answer: (A)

$-\sqrt{25},\sqrt{25}$

The value of $3\cos x+5\sin (x-\pi /6)$
$=3\cos x+5(\sin x\cos \frac{\pi }{6}-\cos x\sin \frac{\pi }{6})$
$=3\cos x+\frac{5\sqrt{3}}{2}\sin x-\frac{1}{2}\cos x$
$=\frac{5}{2}\cos x+\frac{5\sqrt{3}}{2}\sin x$
As $-\sqrt{a^2+b^2}\le a\sin \theta +b\cos \theta \le \sqrt{a^2+b^2}$
$\therefore -\sqrt{25}\le \frac{5}{2}\cos x+\frac{5\sqrt{3}}{2}\sin x\le \sqrt{25}$
$\Rightarrow -\sqrt{25}\le 3\cos x+5\sin (x-\pi /6)\le \sqrt{25}$