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Use the given value and the trigonometric identities to find the remaining trigonometric functions of the angle.

$\cot \theta = −4$, $\sin \theta > 0$

Then it asks me to find:

$ \sin(\theta) $, $ \cos(\theta) $, $ \tan(\theta) $, $ \csc(\theta) $, and $ \sec(\theta) $

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    $\begingroup$ Do you mean to find the product $\sin\theta\times\cos\theta \ldots$ or to find each of them separately? $\endgroup$ – James Jul 15 '14 at 15:19
  • $\begingroup$ cot θ = −4 means θ is in 2nd or the 4th quadrant. sin θ > 0 means θ is in the 1st quadrant or the 2nd quadrant. The "," mean "AND". $\endgroup$ – Mick Jul 15 '14 at 15:19
  • $\begingroup$ Given that information they are asking me to find the rest of the 5 functions :/ $\endgroup$ – Kelly Jul 15 '14 at 15:21
  • $\begingroup$ Ok, then put commas between them! $\tan\theta$ should be easy when you remember that $(\tan\theta)^{-1} = \cot\theta$, for the others you have to narrow down the correct quadrant like Mick tells you, which will give you the sign of $\sin\theta$, $\cos\theta$ etc, then you need to try to use any other identities that you know involving $\cot$. $\endgroup$ – James Jul 15 '14 at 15:24
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    $\begingroup$ Draw a triangle in the correct quadrant with an angle $\theta$ such that $\cot\theta = -4$ (label the sides appropriately so that the ratio of the proper sides is $-4$). Use that triangle to figure out what the remaining functions are. $\endgroup$ – rogerl Jul 15 '14 at 15:29
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Recall that $(\tan \theta)^{-1}=\cot\theta$, so $\tan\theta=-\dfrac{1}{4}$. Then, since $\cot\theta=-4$ and $\sin\theta>0$, $\theta$ is in the second quadrant. Note that $\cos\theta<0$; in fact, $-4\cos\theta=\sin \theta$. Now, $$\sin^2\theta+\cos^2\theta=1\implies \cos^2\theta+16\cos^2\theta=1\implies\\ \cos^2\theta=\sqrt{\dfrac{1}{17}}$$ $\sqrt{\dfrac{1}{17}}$ is the negative root of $\dfrac{1}{17}$ (in the following, by $\sqrt{\dfrac{1}{17}}$ I will mean the negative root of $\dfrac{1}{17}$), since $\cos\theta<0$. Then, $\sin\theta=-4\sqrt{\dfrac{1}{17}}$. The rest follows.

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Let $adj$ be x axis of a right triangle, $opp$ be the y axis of the right triangle, and $hyp$ be the hypotenuse.

Since $\cot(\theta) = \frac{adj}{opp}$, this means $\cot(\theta) = -4$ means $adj = \pm4$ or $opp = \pm1$. To really know the sign of $adj$ and $opp$, we need to take into account another given equation $\sin(\theta) > 0$. Since $\sin(\theta) = \frac{opp}{hyp}$, it also suggest $opp > 0$ ($hyp$ is always positive anyway). Thus we know $adj = -4$ and $opp = 1$.

Before going, solve for $hyp$ with pythagorean theorem,

$hyp^2 = (-4)^2 + 1^2$

Once you get $hyp$, just plugin $adj, $hype, opp$ to their respective spot,

$$sin\theta = \frac{opp}{hyp}$$

$$tan\theta = \frac{opp}{adj}$$

And so on...

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