Trigonometric ratios in the circumference

A unit circle is a circle that has its center at the origin of coordinates and its radius is $$1$$. The unit circle in the coordinate axes define four quadrants that are numbered in a counter-clockwise.

$$QOP$$ and $$TOS$$ are similar triangles.

$$QOP$$ and $$T'OS'$$ are similar triangles.


The sine is the coordinate in the $$y$$-axis, and the cosine is the coordinate in the $$x$$-axis and, looking at the image, we see that:

$$$ -1\leqslant \sin(\alpha) \leqslant 1 \quad \text{ and } \quad -1\leqslant \cos(\alpha) \leqslant 1 $$$

In addition, we can see that the sine, cosine and tangent of the angle can be found using the following relations:

$$\sin(\alpha)=\dfrac{\overline{PQ}}{\overline{OP}}= \dfrac{\overline{PQ}}{r}=\overline{PQ}$$

$$\cos(\alpha)=\dfrac{\overline{OQ}}{\overline{OP}}= \overline{OQ}$$

$$\tan(\alpha)=\dfrac{\overline{PQ}}{\overline{OQ}}= \dfrac{\overline{ST}}{\overline{OT}}=\overline{ST}$$

And the inverse trigonometric relationships are:

$$\csc(\alpha)=\dfrac{\overline{OP}}{\overline{PQ}}= \dfrac{\overline{OS'}}{\overline{OT}}=\overline{OS'}$$

$$\sec(\alpha)=\dfrac{\overline{OP}}{\overline{OQ}}= \dfrac{\overline{OS}}{\overline{OT}} \overline{OS}$$

$$\cot(\alpha)=\dfrac{\overline{OQ}}{\overline{PQ}}= \dfrac{\overline{S'T'}}{\overline{OT'}}=\overline{S'T'}$$

Sign of trigonometric ratios

Now, we are going to give the signs taken by the sine and cosine in the unit circle:


And at the limits of each quadrant:

$$ \begin{array}{lcccc} \alpha: & 0^\circ & 90^\circ & 180^\circ & 270^\circ \\ \sin(\alpha) & 0 & 1 & 0 & -1 \\ \cos(\alpha) & 1 & 0 & -1 & 0 \\ \tan(\alpha) & 0 & \rightarrow\infty & 0 & \rightarrow-\infty \\ \end{array}$$

Complementary angles

Two angles $$x$$ and $$y$$ are complementary angles if its sum is a right angle. That is,

  • If $$x+y=90^\circ$$ with $$x$$, $$y$$ in sexagesimal degrees.
  • If $$x+y=\dfrac{\pi}{2}$$ with $$x$$, $$y$$ in radians.

In addition, if two complementary angles are adjacents, their non common sides form a right angle. For example, if $$x=30^\circ$$, its complementary is $$y=60^\circ$$, since $$x+y=30+60=90^\circ$$.

In the following picture, we will see the relationships that appear between the sine and the cosine:


Then, we see that the following relationships are satisfied:




In this example, let's calculate the basic trigonometric ratios of the following angles:

a) $$150^\circ$$:

  • $$\sin(150^\circ) = \cos(90^\circ - 150^\circ) = \cos(-60^\circ) = \cos(60^\circ)= \dfrac{1}{2}$$ (since it is an even function, $$\cos(-\alpha)=\cos(\alpha)$$)

  • $$\cos(150^\circ) = \sin(-60^\circ) = -\sin(60^\circ)= - \dfrac{\sqrt{3}}{2}$$ (since it is an odd function, $$\sin(-\alpha)=-\sin(\alpha)$$)

  • $$\tan(150^\circ) = \dfrac{\dfrac{1}{2}}{- \dfrac{\sqrt{3}}{2}}= -\dfrac{\sqrt{3}}{3}$$

b) $$330^\circ$$:

  • $$\sin(330^\circ) = \sin(-30^\circ) = - \sin(30^\circ) = -\dfrac{1}{2}$$

  • $$\cos(330^\circ) = \cos(-30^\circ) = \cos(30^\circ) = \dfrac{\sqrt{3}}{2}$$

  • $$\tan(330^\circ) = \dfrac{\sin(330^\circ)}{\cos(330^\circ)}= \dfrac{-\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}} =-\dfrac{\sqrt{3}}{3}$$