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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 $\cot.$

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2 Answers 2

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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.

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  • $\begingroup$ Thank you for the response! So $\sin\alpha<0$ for every $\alpha\in(180^\circ,360^\circ)?$ $\endgroup$
    – Katherine
    Dec 30, 2020 at 15:14
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    $\begingroup$ 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 $y=0$ $\endgroup$ Dec 30, 2020 at 15:30
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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 :)

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