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:

$$\sin(\dfrac{\pi}{2}-\alpha)=\cos(\alpha)$$

$$\cos(\dfrac{\pi}{2}-\alpha)=\sin(\alpha)$$

$$\tan(\dfrac{\pi}{2}-\alpha)=\cot(\alpha)$$

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

a) $120^\circ$:

• $\sin(120^\circ) = \cos(90^\circ - 120^\circ) = \cos(-30^\circ) = \cos(30^\circ)= \dfrac{\sqrt{3}}{2}$ (since it is an even function, $\cos(-\alpha)=\cos(\alpha)$)

• $\cos(120^\circ) = \sin(-30^\circ) = -\sin(30^\circ)= - \dfrac{1}{2}$ (since it is an odd function, $\sin(-\alpha)=-\sin(\alpha)$)

• $\tan(120^\circ) = \dfrac{\dfrac{\sqrt{3}}{2}}{- \dfrac{1}{2}}= -\sqrt{3}$

b) $300^\circ$:

• $\sin(300^\circ) = \sin(-60^\circ) = - \sin(60^\circ) = -\dfrac{\sqrt{3}}{2}$

• $\cos(300^\circ) = \cos(-60^\circ) = \cos(60^\circ) = \dfrac{1}{2}$

• $\tan(300^\circ) = \dfrac{\sin(300^\circ)}{\cos(300^\circ)}= \dfrac{-\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} =-\sqrt{3}$