# Find $\sin\alpha,\cos\alpha,\tan\alpha$ if $\cot\alpha=-2$

Find $$\sin\alpha,\cos\alpha,\tan\alpha$$ if $$\cot\alpha=-2.$$

We have defined trigonometry with a circle, but only for angles between $$0^\circ$$ and $$180^\circ.$$

We have $$\begin{cases}\cot\alpha=-2\\\sin^2\alpha+\cos^2\alpha=1\end{cases}\iff\begin{cases}\dfrac{\cos\alpha}{\sin\alpha}=-2\\\sin^2\alpha+\cos^2\alpha=1\end{cases}.$$

So I got that $$\sin\alpha=\dfrac{\sqrt5}{5},\cos\alpha=-\dfrac{2\sqrt5}{5},\tan\alpha=-\dfrac{1}{2},\cot\alpha=-2$$ or $$\sin\alpha=-\dfrac{\sqrt5}{5},\cos\alpha=\dfrac{2\sqrt5}{5},\tan\alpha=-\dfrac{1}{2},\cot\alpha=-2.$$

An angle with sine equal to $$-\dfrac{\sqrt5}{5}$$ isn't in the inverval $$\left[0^\circ;180^\circ\right],$$ right? I suppose that it is possible two different angles to have equal $$\tan$$ and $$\cot.$$

## 2 Answers

There is no angle in the range $$[0,180^\circ]$$ with a negative sine. The angles in $$(180^\circ,360^\circ)$$ all have negative sine. You can use the relation $$\sin(180^\circ+x)=-\sin(x)$$ but if your definition is restricted as you say, only your first answer is acceptable. There are two angles with the same $$\tan$$ and $$\cot$$ in $$[0,360^\circ)$$. They are $$180^\circ$$ apart.

• Thank you for the response! So $\sin\alpha<0$ for every $\alpha\in(180^\circ,360^\circ)?$ Dec 30, 2020 at 15:14
• Yes, that is correct. In the unit circle, $\sin \alpha$ is the projection of the point on the $y$ axis. All the points in that range are below $y=0$ Dec 30, 2020 at 15:30

Yes, more than two angles can have same $$\tan$$ (and $$\cot$$ which solely depends on $$\tan$$). For example, for $$\tan\theta=1$$, you have $$\theta=\frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4},\cdots$$ For any interval of length $$2\pi$$, there are two such angles. For $$\tan\theta=1$$ in $$[0,2\pi]$$, we have $$\theta=\frac{\pi}{4}\textrm{ or }\frac{5\pi}{4}$$ For any interval of length $$\pi$$, you can find only one unique angle. For example, consider your equation. There is only one solution in $$[0,\pi]$$, $$\alpha=\text{arccot}\,(-2)$$

Hope this helps. Ask anything if not clear :)