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Trigonometric Functions — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Functions5 Marks
Question
Prove that $(2\sqrt{3}+3)\sin\text{x}+2\sqrt{3}\cos\text{x}$ lies between $-(2\sqrt{3}+\sqrt{15})$ and $(2\sqrt{3}+\sqrt{15}).$
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Answer

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

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