if $\tan \theta = \sqrt{63}$ and $\cos \theta$ is negative, find $\sin \theta$.
So since $\tan \theta$ is positive and $\cos \theta$ is negative, it lies in the $3$rd quadrant. So $\sin$ is negative, but I don't know how to find $\sin \theta$, please guide me... Thank you.
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$\begingroup$HINT
Remember $\tan \theta$ is just $\frac{\sin \theta}{\cos \theta}$. Here you have $\tan \theta = \frac{\sqrt{63}}{1}$ Now, draw a triangle with the sides as $\sqrt{63}$ and $1$. Now you should be able to find $\sin \theta$ and adjust the signs.
FURTHER HINT
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In general $$\tan^2A=\frac{\sin^2A}{\cos^2A}=\sin^2A\cdot\sec^2A\iff\sin^2A=\frac{\tan^2A}{\sec^2A}=\frac{\tan^2A}{1+\tan^2A}$$
$\endgroup$ 4 $\begingroup$$\tan \theta = \sqrt{63} =\frac {\sqrt{63}}{1} = \frac {y-ordinate}{x-ordinate} =\frac {-\sqrt{63}}{-1}$; since it lies in the 3rd quadrant.
Then, $\sin \theta = \frac {y-ordinate}{radius} = \frac {-\sqrt{63}}{8}$
$\sqrt (x-ordinate^2 + y-ordinate^2)$ = 8 = radius, which is always positive.
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