98.0k views
3 votes
Given coso= 4/9 and csc 0<0, find sin0 and tan0

Given coso= 4/9 and csc 0<0, find sin0 and tan0-example-1
User Jadeja RJ
by
5.9k points

1 Answer

2 votes

Answer:


\sin\theta=-(√(65))/(9).


\tan\theta=-(√(65))/(4).

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((4)/(9)\right)^2,\\ \\\sin^2\theta=1-(16)/(81),\\ \\\sin^2\theta=(65)/(81).

2. Since
\csc\theta=(1)/(\sin\theta)<0, you can state that
\sin\theta<0 and


\sin\theta=-\sqrt{(65)/(81)}=-(√(65))/(9).

3. Use the definition:


\tan\theta=(\sin\theta)/(\cos\theta)=-((√(65))/(9))/((4)/(9))=-(√(65))/(4).

User Naveen Kumar Alone
by
5.4k points