Question

4pi/3 find the exact trigonometric ratios for the angle whose radian measue is given

Answer 1

Answer:

All trigonometric ratios.

Step-by-step explanation:

We have to find all the trigonometric ratios.

Given:

$\theta = \dfrac{4\pi}{3} = \pi + \dfrac{\pi}{3}$

Formula:

$\sin(\pi +x) = -\sin(x)\\\cos(\pi +x) = -\cos(x)$

$\sin(\pi + \dfrac{\pi}{3}) = -\sin(\dfrac{\pi}{3}) = -\dfrac{\sqrt{3}}{2}$

$\cos ((\pi + \dfrac{\pi}{3})) = -\cos(\dfrac{\pi}{3}) = -\dfrac{1}{2}$

Formula:

$\tan\theta =\dfrac{\sin\theta}{\cos\theta}\\\cot\theta = \dfrac{1}{\tan\theta}\\\sec\theta = \dfrac{1}{\cos\theta}\\\csc\theta = \dfrac{1}{\sin\theta}$

$\tan(\pi + \dfrac{\pi}{3})= \dfrac{\sin(\pi + \dfrac{\pi}{3})}{\cos(\pi + \dfrac{\pi}{3})}\\\\\tan(\pi + \dfrac{\pi}{3}) = \dfrac{\frac{-\sqrt{3}}{2}}{\frac{-1}{2}} = \sqrt{3}$

$\cot(\pi + \dfrac{\pi}{3}) = \dfrac{1}{\tan(\pi + \dfrac{\pi}{3})} = \dfrac{1}{\sqrt{3}}$

$\sec(\pi + \dfrac{\pi}{3}) = \dfrac{1}{\cos(\pi + \dfrac{\pi}{3})} = -2$

$\csc(\pi + \dfrac{\pi}{3}) = \dfrac{1}{\sin(\pi + \dfrac{\pi}{3})} = -\dfrac{2}{\sqrt{3}}$