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.



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.


A triangle

  • $\begingroup$ why is cosθ = 1? $\endgroup$ – user2463158 Jun 23 '14 at 12:36
  • $\begingroup$ @user2463158 $\cos \theta$ is not equal to $1$. Find the hypotenuse then we have $\cos \theta = \frac{\text{adj.}}{\text{hyp.}}$ What do you get? $\endgroup$ – Jeel Shah Jun 23 '14 at 12:39
  • $\begingroup$ So hypo is 8 and answer is sin=opp/hypo=1/8 ?? $\endgroup$ – user2463158 Jun 23 '14 at 12:51
  • $\begingroup$ Nope. Think again :), you accidentally calculated $\cos$ :). Look at where our reference angle is and then look at the side directly across it, which number do you get? Also, remember, $\sin \theta$ has to negative, then which one will negative? The opposite side or the hypotenuse? $\endgroup$ – Jeel Shah Jun 23 '14 at 12:52
  • $\begingroup$ so my answer is sinθ= -√63/8 $\endgroup$ – user2463158 Jun 23 '14 at 13:02

In general $$\tan^2A=\frac{\sin^2A}{\cos^2A}=\sin^2A\cdot\sec^2A\iff\sin^2A=\frac{\tan^2A}{\sec^2A}=\frac{\tan^2A}{1+\tan^2A}$$

  • $\begingroup$ and if $\cos$ is negative and $\tan$ positive, $\sin$ is negative $\endgroup$ – John Fernley Jun 23 '14 at 12:57
  • $\begingroup$ @JohnFernley, He has already derived that $\endgroup$ – lab bhattacharjee Jun 23 '14 at 13:20
  • $\begingroup$ I didn't really understand what the third quadrant meant but I'll take your word for it $\endgroup$ – John Fernley Jun 23 '14 at 13:36
  • $\begingroup$ Thanks for letting me know by the way, you never know what could have happened if we accidentally gave a complete answer $\endgroup$ – John Fernley Jun 23 '14 at 13:37

$\tan \theta = \sqrt{63} =\frac {\sqrt{63}}{1} = \frac {y-ordinate}{x-ordinate} =\frac {-\sqrt{63}}{-1}$; 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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