Question

The maximum value of $3\cos \theta + 5\sin \left( {\theta - \dfrac{\pi }{6}} \right)$ for any real value of $\theta$ is:
A $\dfrac{{\sqrt {79} }}{2}$
B $\sqrt {31}$
C $\sqrt {19}$
D $\sqrt {34}$

Answer 1

In this question we need to find out the value of $3\cos \theta + 5\sin \left( {\theta - \dfrac{\pi }{6}} \right)$, for that firstly we need to use the formula:$\sin (A - B) = \sin A\cos B - \cos A\sin B$. After that we will put the values and as we need to find the maximum value, so for that we will be using the formula:$\sqrt {{a^2} + {b^2}}$.

Complete step by step answer:
We have been provided that we need to find the value of $3\cos \theta + 5\sin \left( {\theta - \dfrac{\pi }{6}} \right)$,
So, firstly we will be using the formula: $\sin (A - B) = \sin A\cos B - \cos A\sin B$ for simplification,
Now the equation becomes: $3\cos \theta + 5\left( {\sin \theta .\cos \left( {\dfrac{\pi }{6}} \right) - \cos \theta .\sin \left( {\dfrac{\pi }{6}} \right)} \right)$,
As we know the value of $\cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2}$ and $\sin \dfrac{\pi }{6} = \dfrac{1}{2}$,
Now put these values in $3\cos \theta + 5\left( {\sin \theta .\cos \left( {\dfrac{\pi }{6}} \right) - \cos \theta .\sin \left( {\dfrac{\pi }{6}} \right)} \right)$,
So, the equation becomes: $3\cos \theta + 5\left( {\sin \theta .\dfrac{{\sqrt 3 }}{2} - \cos \theta .\dfrac{1}{2}} \right)$,
Now open the brackets, and write the terms of $\sin \theta$ and $\cos \theta$ separately,
This will result in: $\left( {\dfrac{{5\sqrt 3 }}{2}} \right)\sin \theta + \left( {3 - \dfrac{5}{2}} \right)\cos \theta$,
Simplifying the above equation, we will get: $\left( {\dfrac{{5\sqrt 3 }}{2}} \right)\sin \theta + \dfrac{1}{2}\cos \theta$,
Now we have got the equation in the form:$a\cos \theta + b\sin \theta$,
And for finding the maximum value we will be using the formula: $\sqrt {{a^2} + {b^2}}$,
So, the maximum value for $3\cos \theta + 5\sin \left( {\theta - \dfrac{\pi }{6}} \right)$:$\sqrt {{{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{{5\sqrt 3 }}{2}} \right)}^2}}$,
Simplifying the above values:$\sqrt {\dfrac{{75 + 1}}{4}} = \sqrt {19}$.
So, the maximum value comes out to be:$\sqrt {19}$.

So, the correct answer is “Option C”.

Note: In this question do not get confused with the formula of maximum and minimum value as both the formulas are almost the same. For finding maximum value the formula used is:$\sqrt {{a^2} + {b^2}}$and that for finding minimum value is: $- \sqrt {{a^2} + {b^2}}$, so be careful with this.