Question

Find the maximum value of $5\cos \theta + 3\cos \left( {\theta + \dfrac{\pi }{3}} \right) + 3$.
A. 5
B. 11
C. 10
D. -11

Answer 1

Hint: First we will apply the formula sum of cosine formula to simplify the given expression. Then apply the formula to find the maximum and minimum of the given expression.

Formula Used:
$\cos \left( {x + y} \right) = \cos x\cos y - \sin x\sin y$
Distributive property $a\left( {b + c} \right) = a \cdot b + a \cdot c$
$ - \sqrt {{a^2} + {b^2}} \le a\cos \theta + b\sin \theta \le \sqrt {{a^2} + {b^2}} $

Complete step by step solution:
Given expression is
$5\cos \theta + 3\cos \left( {\theta + \dfrac{\pi }{3}} \right) + 3$
Now we will apply the formula $\cos \left( {x + y} \right) = \cos x\cos y - \sin x\sin y$ in $\cos \left( {\theta + \dfrac{\pi }{3}} \right)$
$5\cos \theta + 3\cos \left( {\theta + \dfrac{\pi }{3}} \right) + 3 = 5\cos \theta + 3\left[ {\cos \theta \cos \dfrac{\pi }{3} - \sin \theta \sin \dfrac{\pi }{3}} \right] + 3$
Now putting the values $\cos \dfrac{\pi }{3} = \dfrac{1}{2}$ and $\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}$
$5\cos \theta + 3\cos \left( {\theta + \dfrac{\pi }{3}} \right) + 3 = 5\cos \theta + 3\left[ {\dfrac{1}{2}\cos \theta - \dfrac{{\sqrt 3 }}{2}\sin \theta } \right] + 3$
Now we will apply distributive property $a\left( {b + c} \right) = a \cdot b + a \cdot c$ in $3\left[ {\dfrac{1}{2}\cos \theta - \dfrac{{\sqrt 3 }}{2}\sin \theta } \right]$
$5\cos \theta + 3\cos \left( {\theta + \dfrac{\pi }{3}} \right) + 3 = 5\cos \theta + \dfrac{3}{2}\cos \theta - \dfrac{{3\sqrt 3 }}{2}\sin \theta + 3$
Add the like terms
$5\cos \theta + 3\cos \left( {\theta + \dfrac{\pi }{3}} \right) + 3 = \left( {5 + \dfrac{3}{2}} \right)\cos \theta - \dfrac{{3\sqrt 3 }}{2}\sin \theta + 3$
$5\cos \theta + 3\cos \left( {\theta + \dfrac{\pi }{3}} \right) + 3 = \dfrac{{13}}{2}\cos \theta - \dfrac{{3\sqrt 3 }}{2}\sin \theta + 3$
Now compare the expression $\dfrac{{13}}{2}\cos \theta - \dfrac{{3\sqrt 3 }}{2}\sin \theta + 3$ with $a\cos \theta + b\sin \theta + c$
Here $a = \dfrac{{13}}{2}$
$b = - \dfrac{{3\sqrt 3 }}{2}$
$c = 3$
We will apply $ - \sqrt {{a^2} + {b^2}} \le a\cos \theta + b\sin \theta \le \sqrt {{a^2} + {b^2}} $ in $\dfrac{{13}}{2}\cos \theta - \dfrac{{3\sqrt 3 }}{2}\sin \theta + 3$
$ - \sqrt {{{\left( {\dfrac{{13}}{2}} \right)}^2} + {{\left( { - \dfrac{{3\sqrt 3 }}{2}} \right)}^2}} \le \dfrac{{13}}{2}\cos \theta - \dfrac{{3\sqrt 3 }}{2}\sin \theta \le \sqrt {{{\left( {\dfrac{{13}}{2}} \right)}^2} + {{\left( { - \dfrac{{3\sqrt 3 }}{2}} \right)}^2}} $
$ \Rightarrow - \sqrt {\dfrac{{169}}{4} + \dfrac{{27}}{4}} \le \dfrac{{13}}{2}\cos \theta - \dfrac{{3\sqrt 3 }}{2}\sin \theta \le \sqrt {\dfrac{{169}}{4} + \dfrac{{27}}{4}} $
$ \Rightarrow - \sqrt {\dfrac{{196}}{4}} \le \dfrac{{13}}{2}\cos \theta - \dfrac{{3\sqrt 3 }}{2}\sin \theta \le \sqrt {\dfrac{{196}}{4}} $
$ \Rightarrow - \sqrt {49} \le \dfrac{{13}}{2}\cos \theta - \dfrac{{3\sqrt 3 }}{2}\sin \theta \le \sqrt {49} $
$ \Rightarrow - 7 \le \dfrac{{13}}{2}\cos \theta - \dfrac{{3\sqrt 3 }}{2}\sin \theta \le 7$
Add 3 on both sides of inequality
$ - 7 + 3 \le \dfrac{{13}}{2}\cos \theta - \dfrac{{3\sqrt 3 }}{2}\sin \theta + 3 \le 7 + 3$
$ \Rightarrow - 4 \le \dfrac{{13}}{2}\cos \theta - \dfrac{{3\sqrt 3 }}{2}\sin \theta + 3 \le 10$
The maximum value of $\dfrac{{13}}{2}\cos \theta - \dfrac{{3\sqrt 3 }}{2}\sin \theta + 3$ is 10.

Option ‘C’ is correct

Note: Students frequently use the incorrect formula. Instead of using the formula $ - \sqrt {{a^2} + {b^2}} \le a\cos \theta + b\sin \theta \le \sqrt {{a^2} + {b^2}} $, they use $ - \sqrt {{a^2} + {b^2}} \le a\cos \theta + b\sin \theta + c \le \sqrt {{a^2} + {b^2}} $. To calculate the maximum value of the $5\cos \theta + 3\cos \left( {\theta + \dfrac{\pi }{3}} \right) + 3$ use the formula $ - \sqrt {{a^2} + {b^2}} + c \le a\cos \theta + b\sin \theta + c \le \sqrt {{a^2} + {b^2}} + c$