Question

Minimum value of the expression $4\sin (x-\frac{\pi }{6})-6\sqrt{2}\cos (x+\frac{\pi }{4})$ is

• A. $-\sqrt{112+24\sqrt{3}}$
• B. $-\sqrt{28+24\sqrt{3}}$
• C. $-\sqrt{56+2\sqrt{3}}$
• D. $-\sqrt{84+2\sqrt{3}}$

Answer 1

Answer: (A)

$-\sqrt{112+24\sqrt{3}}$

$\sin (x-\frac{\pi }{6})=\sin x\cos \frac{\pi }{6}-\cos x\sin \frac{\pi }{6}$
$=\frac{\sqrt{3}}{2}\sin x-\frac{1}{2}\cos x$
$\cos (x+\frac{\pi }{4})=\cos x\cos \frac{\pi }{4}-\sin x\sin \frac{\pi }{4}$
$=\frac{1}{\sqrt{2}}\cos x-\frac{1}{\sqrt{2}}\sin x$

Now, $4\sin (x-\frac{\pi }{6})-6\sqrt{2}\cos (x+\frac{\pi }{4})$
$=(2\sqrt{3}+6)\sin x-8\cos x$

Minimim value is $-\sqrt{(2\sqrt{3}+6{)}^2+(-8{)}^2}=-\sqrt{112+24\sqrt{3}}$