Question

The maximum value of $5\cos \left(\mathrm{\theta }\right)+3\cos \left(\mathrm{\theta }+\frac{\mathrm{\pi }}{3}\right)+3$ is

• A. $5$
• B. $11$
• C. $10$
• D. $-11$

Answer 1

The correct option is C

$10$

Explanation for the correct option:

$$\begin{gathered} 5\cos \left(\mathrm{\theta }\right)+3\cos \left(\mathrm{\theta }+\frac{\mathrm{\pi }}{3}\right)+3 \\ =5\cos \left(\mathrm{\theta }\right)+3\left[\cos \left(\mathrm{\theta }\right)\cos \left(\frac{\mathrm{\pi }}{3}\right)–\sin \left(\mathrm{\theta }\right)\sin \left(\frac{\mathrm{\pi }}{3}\right)\right]+3 \\ =5\cos \left(\mathrm{\theta }\right)+3\left[\right(1/2)\cos \left(\mathrm{\theta }\right)–(\surd 3/2)\sin \left(\mathrm{\theta }\right)]+3 \\ =(5+3/2)\cos \left(\mathrm{\theta }\right)–(3\surd 3/2)\sin \left(\mathrm{\theta }\right)+3 \\ =(13/2)\cos \left(\mathrm{\theta }\right)–(3\surd 3/2)\sin \left(\mathrm{\theta }\right)+3 \end{gathered}$$

We know that the maximum value of $a\cos \left(\theta \right)+b\sin \left(\theta \right)+c$ is $c+\sqrt{(a^2+b^2)}$.

Substituting $a=\frac{13}{2}$, $b=-\frac{3\surd 3}{2}$ and $c=3$, we get,

The maximum value $=3+\sqrt{\frac{169}{4}+\frac{27}{4}}=3+7=10$

Hence, $\mathrm{Option}\left(\mathrm{C}\right)$ is the correct answer.