Question:
Prove that $(2 \sqrt{3}+3) \sin x+2 \sqrt{3} \cos x$ lies between $-(2 \sqrt{3}+\sqrt{15})$ and $(2 \sqrt{3}+\sqrt{15})$.
Solution:
Let $f(x)=(2 \sqrt{3}+3) \sin x+2 \sqrt{3} \cos x$
We know that,
$-\sqrt{(2 \sqrt{3}+3)^{2}+(2 \sqrt{3})^{2}} \leq f(x) \leq \sqrt{(2 \sqrt{3}+3)^{2}+(2 \sqrt{3})^{2}}$
$\Rightarrow-\sqrt{12+9+12 \sqrt{3}+12} \leq f(x) \leq \sqrt{12+9+12 \sqrt{3}+12}$
$\Rightarrow-\sqrt{33+12 \sqrt{3}} \leq f(x) \leq \sqrt{33+12 \sqrt{3}}$
Disclaimer : Instead of $-(2 \sqrt{3}+\sqrt{15})$ and $(2 \sqrt{3}+\sqrt{15})$, it should be $-\sqrt{33+12 \sqrt{3}}$ and $\sqrt{33+12 \sqrt{3}}$.