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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.

Vinod Kumar Punia
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2 Answers2

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$$2\sqrt{11+4\sqrt{3}} \approx 8.46834180469$$

$$2(2+\sqrt(5))\approx 8.472135955$$

$$2(2+\sqrt(5)) >2\sqrt{11+4\sqrt{3}}$$

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

Chill2Macht
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You want to show $\sqrt{11+4\sqrt3} < 2 + \sqrt5 \iff 11+4\sqrt3 < 9+4\sqrt5 \iff 1 < 2(\sqrt5-\sqrt3) \\ \iff 1 < 4 (8-2\sqrt{15}) \iff 8\sqrt{15} < 31 \iff 960 < 961$

Macavity
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