if $\tan \theta = \sqrt63$ and also $\cos \theta$ is negative, uncover $\sin \theta$.

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So because $\tan \theta$ is positive and $\cos \theta$ is negative, the lies in the $3$rd quadrant. So $\sin$ is negative, however I don"t know how to uncover $\sin \theta$, please overview me... Say thanks to you.  HINT

Remember $\tan \theta$ is just $\frac\sin \theta\cos \theta$. Below you have actually $\tan \theta = \frac\sqrt631$ Now, draw a triangle through the sides as $\sqrt63$ and also $1$. Now you should be able to find $\sin \theta$ and readjust the signs.

FURTHER HINT  In general $$\tan^2A=\frac\sin^2A\cos^2A=\sin^2A\cdot\sec^2A\iff\sin^2A=\frac\tan^2A\sec^2A=\frac\tan^2A1+\tan^2A$$ $\tan \theta = \sqrt63 =\frac \sqrt631 = \frac y-ordinatex-ordinate =\frac -\sqrt63-1$; since it lies in the 3rd quadrant.

Then, $\sin \theta = \frac y-ordinateradius = \frac -\sqrt638$

$\sqrt (x-ordinate^2 + y-ordinate^2)$ = 8 = radius, which is constantly positive. Thanks because that contributing an answer to thedesigningfairy.comematics ridge Exchange!

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Please aid with mine Trigonometry question $A-B\equiv (2\theta - \tan\theta - \sin\theta \cos\theta)$
when finding $\sin\theta$ and also $\cos\theta$ provided $\tan\theta = \frac35$, shouldn't I gain positive and an unfavorable answers?