Question

Given coso= 4/9 and csc 0<0, find sin0 and tan0

Answer 1

Answer:

$\sin\theta=-\dfrac{\sqrt{65}}{9}.$

$\tan\theta=-\dfrac{\sqrt{65}}{4}.$

Step-by-step explanation:

1. Use the man trigonometric equality

$\cos^2\theta+\sin^2\theta=1.$

From this equality

$\sin^2\theta=1-\cos^2\theta,\\ \\\sin^2\theta=1-\left(\dfrac{4}{9}\right)^2,\\ \\\sin^2\theta=1-\dfrac{16}{81},\\ \\\sin^2\theta=\dfrac{65}{81}.$

2. Since $\csc\theta=\dfrac{1}{\sin\theta}$ you can state that $\sin\theta$ and

$\sin\theta=-\sqrt{\dfrac{65}{81}}=-\dfrac{\sqrt{65}}{9}.$

3. Use the definition:

$\tan\theta=\dfrac{\sin\theta}{\cos\theta}=-\dfrac{\frac{\sqrt{65}}{9}}{\frac{4}{9}}=-\dfrac{\sqrt{65}}{4}.$