Trigonometric Functions on a unit circle

I have to find all solutions for $\theta$ in the given range: $$tan (\theta) = \frac {-1}{\sqrt3}, -\pi \le \theta \lt 2\pi$$

I said that if $(x,y)$ is on the unit circle we have $$\frac{y}{x} = \frac{-1}{\sqrt3}$$ since $x^2+y^2=1$ $\implies x = \frac{\sqrt 3}{2}$,$y=-\frac{1}{2}$ and $x = - \frac{\sqrt3}{2}$,$y=\frac{1}{2}$

I am struggling on how to find the angles because I didn't understand the concept. I put my points into the circle but then I get confused about the angles. The one angle it will be $\theta = -\frac{\pi}{6}$ since $tan^{-1}(-\frac{1}{\sqrt3})= - \frac{\pi}{6}$.

Can anyone help me to understand the way of thinking for the other angles?

With Thanks

• You made an error just above your graph. You meant $$\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}$$ – N. F. Taussig Oct 30 '14 at 10:37

As you determined,

$$\theta = \arctan\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}$$

is one solution.

Remember that the tangent function has period $\pi$. Thus,

$$\tan\theta = -\frac{1}{\sqrt{3}}$$

when

$$\theta = -\frac{\pi}{6} + n\pi, n \in \mathbb{Z}$$

where $\mathbb{Z}$ is the set of integers. Since you require that $-\pi \leq \theta < 2\pi$, you can find the remaining solutions in the interval by finding values of $n$ such that

$$-\pi \leq -\frac{\pi}{6} + n\pi < 2\pi$$

• In case we have sin or cos instead of tan? – user161260 Oct 30 '14 at 10:43
• For sine, you would find a particular solution $\theta$. Since $\sin(\pi - \theta) = \sin\theta$, $\pi - \theta$ is another solution. To find additional solutions, add integer multiples of $2\pi$ to $\theta$ and $\pi - \theta$ to find solutions in the right range. For cosine, after finding a particular solution $\theta$, you use the fact that $\cos(-\theta) = \cos\theta$ to find another solution. Since cosine also has period $2\pi$, you would add integer multiples of $2\pi$ to $\theta$ and $-\theta$ to find values in the right range. – N. F. Taussig Oct 30 '14 at 10:52
• For the example with the tan the solutions will be: $\theta = -\frac{\pi}{6}, \frac{5 \pi}{6} and \frac{11 \pi} {6}$ ? – user161260 Oct 30 '14 at 11:17
• That is correct. – N. F. Taussig Oct 30 '14 at 12:01