Question

Determine which quadrant the angle Ф lies and the reference angle.
Then, find the indicated ratio for Ф' (must be an exact value).
Finally, determine the value for the original expression using the quadrants.

csc$\cfrac{4\pi }{3}$

Answer 1

Answer:

$\csc \left(\dfrac{4\pi}{3}\right)=-\dfrac{2}{\sqrt{3}}$

Step-by-step explanation:

$\phi=\dfrac{4 \pi}{3}$

Therefore, this angle lies in quadrant III since it is between π and 3π/2

It's reference angle is:

$\dfrac{4 \pi}{3}-\pi=\dfrac{\pi}{3}$

and so lies in quadrant I

$\csc (\phi)=\dfrac{1}{\sin(\phi)}$

For sine, quadrant I and II are positive and quadrant III and IV are negative.

Therefore, as  $\sin \dfrac{\pi}{3}=\dfrac{\sqrt{3}}{2}$  then $\sin \dfrac{4\pi}{3}=-\dfrac{\sqrt{3}}{2}$

Finally,

$\csc \left(\dfrac{4\pi}{3}\right)=\dfrac{1}{\sin\left(\dfrac{4\pi}{3}\right)}$

$\implies \csc \left(\dfrac{4\pi}{3}\right)=\dfrac{1}{-\dfrac{\sqrt{3}}{2}}$

$\implies \csc \left(\dfrac{4\pi}{3}\right)=-\dfrac{2}{\sqrt{3}}$

Answer 2

$\\ \rm\Rrightarrow csc\dfrac{4\pi}{3}$

$\\ \rm\Rrightarrow csc\left(\pi +\dfrac{\pi}{3}\right)$

• So it lies in Q3 and it's reference angle is π/3
• Reference angle like usual lies in Q1 .

Now for value

$\\ \rm\Rrightarrow csc\left(\pi+\dfrac{\pi}{3}\right)$

• In Q1 All are positive
• In Q2 sine and cosec are positive .
• In. Q3 tan and cot are positive .
• In Q4 cos and sec are positive.

As it lies in Q3 it's negative

$\\ \rm\Rrightarrow csc\left(-\dfrac{\pi}{3}\right)$

$\\ \rm\Rrightarrow -csc\dfrac{\pi}{3}$

$\\ \rm\Rrightarrow -\dfrac{2}{\sqrt{3}}$