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 $T^{\prime }O{S}^{\prime }$ 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:

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

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

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

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

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

And the inverse trigonometric relationships are:

$\csc (\alpha )={\displaystyle \frac{\overline{OP}}{\overline{PQ}}}={\displaystyle \frac{\overline{O{S}^{\prime }}}{\overline{OT}}}=\overline{O{S}^{\prime }}$

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

$\cot (\alpha )={\displaystyle \frac{\overline{OQ}}{\overline{PQ}}}={\displaystyle \frac{\overline{S^{\prime }{T}^{\prime }}}{\overline{O{T}^{\prime }}}}=\overline{S^{\prime }{T}^{\prime }}$

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:

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

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 $x+y={\displaystyle \frac{\pi }{2}}$ 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:

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

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

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