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 $\cot.$
Best Answer
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.