Sine and cosine on the unit circle | Trigonometry

The unit circle

Put an angle $\theta$ in standard position. Its terminal side crosses the unit circle at exactly one point $$P$$ .

Sine and cosine for any angle

This definition works for any angle, not just acute angles in triangles: obtuse, negative, larger than a full turn. Since $$P$$ is on a circle of radius $$1$$ :

-1 \le \sin\theta \le 1 \qquad\qquad -1 \le \cos\theta \le 1

Quadrantal angles

Angles whose terminal side lies on an axis are quadrantal. Their values come straight from the points where the circle meets the axes:

$\theta$	$$0$$	$\frac{\pi}{2}$	$\pi$	$\frac{3\pi}{2}$
point	$$(1, 0)$$	$$(0, 1)$$	$$(-1, 0)$$	$$(0, -1)$$
$\cos\theta$	$$1$$	$$0$$	$$-1$$	$$0$$
$\sin\theta$	$$0$$	$$1$$	$$0$$	$$-1$$

The Pythagorean identity

$P(\cos\theta, \sin\theta)$ lies on $$x^2 + y^2 = 1$$ , so:

Given one of sine and cosine, the identity gives the other up to sign. The quadrant decides the sign.

Signs in each quadrant

Quadrant	$\sin\theta$	$\cos\theta$
I	$$+$$	$$+$$
II	$$+$$	$$-$$
III	$$-$$	$$-$$
IV	$$-$$	$$+$$

Practice

Quick check

What is $\cos\pi$ ?

Quick check

In which quadrant are both sine and cosine negative?

$\theta$	$\(0\)$	$\frac{\pi}{2}$	$\pi$	$\frac{3\pi}{2}$
point	$\((1, 0)\)$	$\((0, 1)\)$	$\((-1, 0)\)$	$\((0, -1)\)$
$\cos\theta$	$\(1\)$	$\(0\)$	$\(-1\)$	$\(0\)$
$\sin\theta$	$\(0\)$	$\(1\)$	$\(0\)$	$\(-1\)$

Quadrant	$\sin\theta$	$\cos\theta$
I	$\(+\)$	$\(+\)$
II	$\(+\)$	$\(-\)$
III	$\(-\)$	$\(-\)$
IV	$\(-\)$	$\(+\)$