<s>[INST] Prove that $(2\sqrt3+4)\sin x+4\cos x$ lies between $-2(2+\sqrt5)$ and $2(2+\sqrt5)$. Prove that $(2\sqrt3+4)\sin x+4\cos x$ lies between $-2(2+\sqrt5)$ and $2(2+\sqrt5)$.

Since we know that the minimum and maximum values of $a\cos x+b\sin x$ is $-\sqrt{a^2+b^2}$ and $\sqrt{a^2+b^2}$
I applied this formula to get the minimum and maximum values of $(2\sqrt3+4)\sin x+4\cos x$ are $-\sqrt{(2\sqrt3+4)^2+(4)^2}$ and $\sqrt{(2\sqrt3+4)^2+(4)^2}$
$\sqrt{(2\sqrt3+4)^2+(4)^2}=\sqrt{12+16+16+16\sqrt3}=\sqrt{44+16\sqrt3}=2\sqrt{11+4\sqrt3}\neq 2(2+\sqrt5)$
I do not know where i am wrong or is there some other method possible?Please help.
 [/INST] You want to show $\sqrt{11+4\sqrt3} < 2 + \sqrt5 \iff 11+4\sqrt3 < 9+4\sqrt5 \iff 1 < 2(\sqrt5-\sqrt3) \\ \iff 1 < 4 (8-2\sqrt{15}) \iff 8\sqrt{15} < 31 \iff 960 < 961$
</s>