Question

Find the value of ${\sin ^3}\left( {1099\pi - \dfrac{\pi }{6}} \right) + {\cos ^3}\left( {50\pi - \dfrac{\pi }{3}} \right)$
A) $\dfrac{1}{4}$
B) 0
C)$\sqrt 3 $
D) None of these

Answer 1

Hint: Remember the given formula for solving these types of questions,
  $
  \sin \left( {2k\pi + \theta } \right) = \sin \theta \\
  \cos \left( {2k\pi - \theta } \right) = \cos \theta {\text{ }};{\text{ }}where{\text{ }}k \in \mathbb{Z} \\
 $
We use these formulae to find the value of $\sin \left( {1099\pi - \dfrac{\pi }{6}} \right)$ and $\cos \left( {50\pi - \dfrac{\pi }{3}} \right)$ respectively, and them cube them and add them to reach the required answer.

 Complete step-by-step answer:
Given, ${\sin ^3}\left( {1099\pi - \dfrac{\pi }{6}} \right) + {\cos ^3}\left( {50\pi - \dfrac{\pi }{3}} \right)$
Note that,
  $
  \sin \left( {2k\pi + \theta } \right) = \sin \theta \\
  \cos \left( {2k\pi - \theta } \right) = \cos \theta {\text{ }};{\text{ }}where{\text{ }}k \in \mathbb{Z} \\
 $
Now,
 ${\sin ^3}\left( {1099\pi - \dfrac{\pi }{6}} \right) + {\cos ^3}\left( {50\pi - \dfrac{\pi }{3}} \right)$ , can be written as,
Now substituting $1099\pi = 2\left( {549} \right)\pi + \pi $
 \[ = {\sin ^3}\left( {2\left( {549} \right)\pi + \pi - \dfrac{\pi }{6}} \right) + {\cos ^3}\left( {2 \times 25 \times \pi - \dfrac{\pi }{3}} \right)\]
As, we have $\sin \left( {2k\pi + \theta } \right) = \sin \theta $ and $\cos \left( {2k\pi - \theta } \right) = \cos \theta {\text{ }}$ , where $k \in \mathbb{Z}$ , so we get,
 $ = {\sin ^3}\left( {\pi - \dfrac{\pi }{6}} \right) + {\cos ^3}\left( {\dfrac{\pi }{3}} \right)$
  $ = {\sin ^3}\left( {\dfrac{{5\pi }}{6}} \right) + {\cos ^3}\left( {\dfrac{\pi }{3}} \right)$
Taking the cube outside we get,
  $ = {\left[ {\sin \left( {\dfrac{{5\pi }}{6}} \right)} \right]^3} + {\left[ {\cos \left( {\dfrac{\pi }{3}} \right)} \right]^3}$
Now as $\sin \left( {\dfrac{{5\pi }}{6}} \right) = \dfrac{1}{2}$ and $\cos \left( {\dfrac{\pi }{3}} \right) = \dfrac{1}{2}$ , we get,
  $ = {\left( {\dfrac{1}{2}} \right)^3} + {\left( {\dfrac{1}{2}} \right)^3}$
On Cubing the terms we get,
  $ = \dfrac{1}{8} + \dfrac{1}{8}$
On simplification we get,
  $ = \dfrac{1}{4}$
Therefore, the value of the value of ${\sin ^3}\left( {1099\pi - \dfrac{\pi }{6}} \right) + {\cos ^3}\left( {50\pi - \dfrac{\pi }{3}} \right)$ is $\dfrac{1}{4}$

∴ Option (A) is correct.

Note: You should remember the basic trigonometric formulas like
$
  \sin \left( {2k\pi + \theta } \right) = \sin \theta \\
  \cos \left( {2k\pi - \theta } \right) = \cos \theta {\text{ }};{\text{ }}where{\text{ }}k \in \mathbb{Z} \\
 $
Also note that always ${\cos ^n}\theta = {\left( {\cos \theta } \right)^n}$is true for any trigonometric ratio and for any value of n.