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$\tan t = \dfrac{1}{4}$, terminal point of $t$ is in the third quadrant.

So $\tan t = \dfrac{-\sqrt{1-\cos^2 t}}{\cos t} = \dfrac{1}{4} $, because it's in the third quadrant. After solving for $\cos^2 t$ get as an answer that $\cos ^2 t = \dfrac{16}{17}$, but apparently this is wrong. Where lies my mistake?

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  • $\begingroup$ What are you trying to find? $\endgroup$ – Ashar Tafhim Sep 11 '16 at 19:48
  • $\begingroup$ @AsharTafhim Values for $\sin t$ and $\cos t$. $\endgroup$ – Nathan Sep 11 '16 at 19:51
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From $\cos^2 t=\frac{16}{17}$ we find $\cos t=\pm\sqrt{\frac{16}{17}}$ and, in a similar way $ \sin t=\pm\sqrt{\frac{1}{17}}$, now, using the fact that $t$ is in the third quadrant we can chose the signs. Can you do this?

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  • $\begingroup$ The sign should be negative, if I'm correct. $\endgroup$ – Nathan Sep 11 '16 at 19:56
  • $\begingroup$ Yes, you are correct! $\endgroup$ – Emilio Novati Sep 11 '16 at 19:57
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If

$$\tan t = \frac{1}{4}$$

then

$$t = \arctan \left(\frac{1}{4}\right) = 14.03°$$

or $0.244$ radians.

Beware

$$\tan t = \frac{\sin t}{\cos t} = \frac{\sqrt{1 - \cos^2t}}{\cos t}$$

In your formula, there is a minus sign which is wrong.

Hence

$$\frac{\sqrt{1 - \cos^2t}}{\cos t} = \frac{1}{4}$$

$$4\sqrt{1 - \cos^2 t} = \cos t$$

Squaring

$$16 - 16\cos^2 t = \cos^2 t$$

$$16 = 17\cos^2 t$$

$$\cos t = \pm \sqrt{\frac{16}{17}} = \pm 4\sqrt{\frac{1}{17}}$$

Then choose the right sign for what you know.

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