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Question: Find the value of \({\sin ^3}\left( {1099\pi - \dfrac{\pi }{6}} \right) + {\cos ^3}\left( {50\pi - \...

Find the value of sin⁡3(1099π−π6)+cos⁡3(50π−π3){\sin ^3}\left( {1099\pi - \dfrac{\pi }{6}} \right) + {\cos ^3}\left( {50\pi - \dfrac{\pi }{3}} \right)
A) 14\dfrac{1}{4}
B) 0
C)3\sqrt 3
D) None of these

Explanation

Solution

Remember the given formula for solving these types of questions,
sin⁡(2kπ+θ)=sin⁡θ cos⁡(2kπ−θ)=cos⁡θ ; where k∈Z  \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⁡(1099π−π6)\sin \left( {1099\pi - \dfrac{\pi }{6}} \right) and cos⁡(50π−π3)\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(1099π−π6)+cos⁡3(50π−π3){\sin ^3}\left( {1099\pi - \dfrac{\pi }{6}} \right) + {\cos ^3}\left( {50\pi - \dfrac{\pi }{3}} \right)
Note that,
sin⁡(2kπ+θ)=sin⁡θ cos⁡(2kπ−θ)=cos⁡θ ; where k∈Z  \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(1099π−π6)+cos⁡3(50π−π3){\sin ^3}\left( {1099\pi - \dfrac{\pi }{6}} \right) + {\cos ^3}\left( {50\pi - \dfrac{\pi }{3}} \right) , can be written as,
Now substituting 1099π=2(549)π+π1099\pi = 2\left( {549} \right)\pi + \pi
=sin⁡3(2(549)π+π−π6)+cos⁡3(2×25×π−π3)= {\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⁡(2kπ+θ)=sin⁡θ\sin \left( {2k\pi + \theta } \right) = \sin \theta and cos⁡(2kπ−θ)=cos⁡θ \cos \left( {2k\pi - \theta } \right) = \cos \theta {\text{ }} , where k∈Zk \in \mathbb{Z} , so we get,
=sin⁡3(π−π6)+cos⁡3(π3)= {\sin ^3}\left( {\pi - \dfrac{\pi }{6}} \right) + {\cos ^3}\left( {\dfrac{\pi }{3}} \right)
=sin⁡3(5π6)+cos⁡3(π3)= {\sin ^3}\left( {\dfrac{{5\pi }}{6}} \right) + {\cos ^3}\left( {\dfrac{\pi }{3}} \right)
Taking the cube outside we get,
=[sin⁡(5π6)]3+[cos⁡(π3)]3= {\left[ {\sin \left( {\dfrac{{5\pi }}{6}} \right)} \right]^3} + {\left[ {\cos \left( {\dfrac{\pi }{3}} \right)} \right]^3}
Now as sin⁡(5π6)=12\sin \left( {\dfrac{{5\pi }}{6}} \right) = \dfrac{1}{2} and cos⁡(π3)=12\cos \left( {\dfrac{\pi }{3}} \right) = \dfrac{1}{2} , we get,
=(12)3+(12)3= {\left( {\dfrac{1}{2}} \right)^3} + {\left( {\dfrac{1}{2}} \right)^3}
On Cubing the terms we get,
=18+18= \dfrac{1}{8} + \dfrac{1}{8}
On simplification we get,
=14= \dfrac{1}{4}
Therefore, the value of the value of sin⁡3(1099π−π6)+cos⁡3(50π−π3){\sin ^3}\left( {1099\pi - \dfrac{\pi }{6}} \right) + {\cos ^3}\left( {50\pi - \dfrac{\pi }{3}} \right) is 14\dfrac{1}{4}

∴ Option (A) is correct.

Note: You should remember the basic trigonometric formulas like
sin⁡(2kπ+θ)=sin⁡θ cos⁡(2kπ−θ)=cos⁡θ ; where k∈Z  \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⁡nθ=(cos⁡θ)n{\cos ^n}\theta = {\left( {\cos \theta } \right)^n}is true for any trigonometric ratio and for any value of n.