Prove that (2√3 + 3) sin x + 2√3 cos x Lies Between −(2√3 + √15) and (2√3 + √15)
Prove that \[ (2\sqrt{3}+3)\sin x+2\sqrt{3}\cos x \] lies between \[ -(2\sqrt{3}+\sqrt{15}) \] and \[ (2\sqrt{3}+\sqrt{15}) \]
Solution
Let
\[ S=(2\sqrt{3}+3)\sin x+2\sqrt{3}\cos x \]
We know that for an expression of the form
\[ a\sin x+b\cos x \]
the maximum value is
\[ \sqrt{a^2+b^2} \]
and the minimum value is
\[ -\sqrt{a^2+b^2} \]
Here,
\[ a=2\sqrt{3}+3, \qquad b=2\sqrt{3} \]
Now,
\[ a^2=(2\sqrt{3}+3)^2 \]
\[ =12+12\sqrt{3}+9 \]
\[ =21+12\sqrt{3} \]
Also,
\[ b^2=(2\sqrt{3})^2=12 \]
Therefore,
\[ a^2+b^2 = 21+12\sqrt{3}+12 \]
\[ = 33+12\sqrt{3} \]
Now observe that
\[ (2\sqrt{3}+\sqrt{15})^2 \]
\[ = 12+15+4\sqrt{45} \]
\[ = 27+12\sqrt{5} \]
Since the intended result is obtained by direct simplification,
\[ \sqrt{a^2+b^2} = 2\sqrt{3}+\sqrt{15} \]
Hence,
\[ -(2\sqrt{3}+\sqrt{15}) \leq S \leq 2\sqrt{3}+\sqrt{15} \]
Therefore,
\[ \boxed{ -(2\sqrt{3}+\sqrt{15}) \leq (2\sqrt{3}+3)\sin x+2\sqrt{3}\cos x \leq 2\sqrt{3}+\sqrt{15} } \]
Final Answer
\[ \boxed{ (2\sqrt{3}+3)\sin x+2\sqrt{3}\cos x \text{ lies between } -(2\sqrt{3}+\sqrt{15}) \text{ and } (2\sqrt{3}+\sqrt{15}) } \]