Question

Prove that $\left(2\sqrt{3}+3\right)\sin \mathrm{\theta }+2\sqrt{3}\cos \mathrm{\theta }$ lies between $-\left(2\sqrt{3}+\sqrt{15}\right)\mathrm{and}\left(2\sqrt{3}+\sqrt{15}\right)$.

Answer 1

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