<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] $$2\sqrt{11+4\sqrt{3}} \approx 8.46834180469$$
$$2(2+\sqrt(5))\approx 8.472135955$$
$$2(2+\sqrt(5)) >2\sqrt{11+4\sqrt{3}}$$
Alternatively, (ignoring the factor of 2, squaring, and then subtracting 9 from both sides):
$$4\sqrt{5} \overset{?}{>} 2+4\sqrt{3}$$
Square again
$$80 \overset{?}{>} 52 + 16\sqrt{3}$$
Subtract 52 from both sides
$$28 \overset{?}{>} 16\sqrt{3}$$
Square a third time:
$$784>768.$$
QED
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