Chapter 13.6, Problem 33PPS

To determine

To find the exact value of $2sin\left(\frac{4\pi }{3}\right)-3\cos \left(\frac{11\pi }{6}\right)$ .

Answer to Problem 33PPS

  $-\frac{\sqrt{3}}{2}$

Explanation of Solution

Given:

  $2sin\left(\frac{4\pi }{3}\right)-3\cos \left(\frac{11\pi }{6}\right)$

Calculation:

Given expression $2sin\left(\frac{4\pi }{3}\right)-3\cos \left(\frac{11\pi }{6}\right),$

  $\sin \left(\frac{4\pi }{3}\right)=\sin \left(\pi +\frac{\pi }{3}\right)$ $=-\sin \left(\frac{\pi }{3}\right)$   [ $\sin$ is negative in 3^rd quadrant]

  $=-\frac{\sqrt{3}}{2}$   [ $\because \sin \left(\frac{\pi }{3}\right)=\frac{\sqrt{3}}{2}$ ]

And $\cos \left(\frac{11\pi }{6}\right)=\cos \left(2\pi -\frac{\pi }{6}\right)$ $=\cos \left(\frac{\pi }{6}\right)$   [ $\cos$ is positive in 4^th quadrant]

  $=\frac{\sqrt{3}}{2}$   [ $\because \cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}$ ]

Now,

  $2sin\left(\frac{4\pi }{3}\right)-3\cos \left(\frac{11\pi }{6}\right)=2\left(\frac{\sqrt{3}}{2}\right)-3\left(\frac{\sqrt{3}}{2}\right)$

  $\begin{array}{l}=\frac{2\sqrt{3}}{2}-\frac{3\sqrt{3}}{2}\\ =\frac{2\sqrt{3}-3\sqrt{3}}{2}\\ =-\frac{\sqrt{3}}{2}\end{array}$

Conclusion:

Therefore, the exact value of $2sin\left(\frac{4\pi }{3}\right)-3\cos \left(\frac{11\pi }{6}\right)$ is $-\frac{\sqrt{3}}{2}$ .