Question

Find the value of ${\sin }^3(1099\pi -\frac{\pi }{6})+{\cos }^3(50\pi -\frac{\pi }{3})$

• A. $\frac{1}{4}$
• B. $0$
• C. $\sqrt{3}$
• D. None of these

Answer 1

The correct option is A $\frac{1}{4}$

As we know that

$\sin (2k\pi +\theta )=\sin \theta ,k\in Z$

$\cos (2k\pi -\theta )=\cos \theta ,k\in Z$

${\sin }^3(1099\pi -\frac{\pi }{6})+{\cos }^3(50\pi -\frac{\pi }{3})={\sin }^3(2\left(549\right)\pi +\pi -\frac{\pi }{6})+{\cos }^3(2\left(25\right)\pi -\frac{\pi }{3})$

$={\sin }^3\frac{5\pi }{6}+{\cos }^3\frac{\pi }{3}$

$=(\frac{1}{2}{)}^3+(\frac{1}{2}{)}^3$

$=\frac{1+1}{8}$

$=\frac{1}{4}$