3.5 Unit Circle and Trigonometric Identities

1. Sine and Cosine on the Unit Circle

Trigonometric functions are defined using a unit circle (radius $r=1$, centered at the origin) on a Cartesian coordinate system.

For a point $P(x,y)$ on the unit circle, where the radius forms an angle $\theta$ with the positive x-axis:

The sine function equals the vertical displacement: $\sin\theta = y$.
The cosine function equals the horizontal displacement: $\cos\theta = x$.

Therefore, any angle $\theta$ corresponds to a point on the circle where $\mathbf{(\cos\theta, \sin\theta) = (x, y)}$.

Quadrant Analysis (ASTC)

Because the unit circle spans four quadrants, the signs of the trigonometric functions depend on the position of the angle.

1st Quadrant ($0^\circ < \theta < 90^\circ$): Sine and cosine are positive.
2nd Quadrant ($90^\circ < \theta < 180^\circ$): Sine is positive, cosine is negative.
3rd Quadrant ($180^\circ < \theta < 270^\circ$): Sine and cosine are negative.
4th Quadrant ($270^\circ < \theta < 360^\circ$): Sine is negative, cosine is positive.

Note: Due to periodicity, multiple angles map to the same point. Adding multiples of a full rotation ($360^\circ k$ or $2k\pi$) yields the same sine and cosine values.

2. Tangent on the Unit Circle

The tangent function is defined as $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$. Geometrically, it can be modeled using a vertical tangent line touching the unit circle at $(1, 0)$.

Extending the terminal side of angle $\theta$ to intersect this vertical tangent line gives a y-coordinate equal to $\tan\theta$.

Properties of the Tangent Function:

Values span all real numbers: $-\infty < \tan\theta < +\infty$.
The function is undefined when the line is parallel to the vertical axis, which occurs at $90^\circ$ and $270^\circ$ (or $\dfrac{\pi}{2} + k\pi$).
Opposite angles have the same tangent value: $\tan(\theta + 180^\circ) = \tan\theta$.

3. Trigonometric Identities

Trigonometric identities connect different trigonometric functions.

Fundamental Pythagorean Identity:

$$\sin^2\theta + \cos^2\theta = 1$$

Double Angle Identities: These identities convert double angles into single-angle expressions.

$\sin 2\theta = 2\sin\theta\cos\theta$

$\cos 2\theta = \cos^2\theta - \sin^2\theta$

$\cos 2\theta = 2\cos^2\theta - 1$

$\cos 2\theta = 1 - 2\sin^2\theta$

$\tan 2\theta = \dfrac{2\tan\theta}{1 - \tan^2\theta}$

EXAMPLE 1

Given $\sin\theta = \dfrac{3}{5}$, find $\cos\theta$, $\tan\theta$, $\sin 2\theta$, $\cos 2\theta$, and $\tan 2\theta$ for the given domains.

Solution: Use the Pythagorean identity to find the remaining ratios.

$$\begin{aligned} \cos^2\theta &= 1 - \sin^2\theta \\ \cos^2\theta &= 1 - \left(\dfrac{3}{5}\right)^2 = 1 - \dfrac{9}{25} = \dfrac{16}{25} \\ \implies \cos\theta &= \pm\dfrac{4}{5} \end{aligned}$$

(a) For an acute domain ($0^\circ < \theta < 90^\circ$): Cosine is positive.

$\cos\theta = \mathbf{\dfrac{4}{5}}$

$\tan\theta = \dfrac{3/5}{4/5} = \mathbf{\dfrac{3}{4}}$

$\sin 2\theta = 2\left(\dfrac{3}{5}\right)\left(\dfrac{4}{5}\right) = \mathbf{\dfrac{24}{25}}$

$\cos 2\theta = \left(\dfrac{4}{5}\right)^2 - \left(\dfrac{3}{5}\right)^2 = \dfrac{16}{25} - \dfrac{9}{25} = \mathbf{\dfrac{7}{25}}$

$\tan 2\theta = \dfrac{2(3/4)}{1 - (3/4)^2} = \dfrac{6/4}{1 - 9/16} = \dfrac{3/2}{7/16} = \mathbf{\dfrac{24}{7}}$

(b) For an obtuse domain ($90^\circ < \theta < 180^\circ$): Cosine is negative.

$\cos\theta = \mathbf{-\dfrac{4}{5}}$

$\tan\theta = \dfrac{3/5}{-4/5} = \mathbf{-\dfrac{3}{4}}$

$\sin 2\theta = 2\left(\dfrac{3}{5}\right)\left(-\dfrac{4}{5}\right) = \mathbf{-\dfrac{24}{25}}$

$\cos 2\theta = \left(-\dfrac{4}{5}\right)^2 - \left(\dfrac{3}{5}\right)^2 = \mathbf{\dfrac{7}{25}}$

$\tan 2\theta = \mathbf{-\dfrac{24}{7}}$

Notice on Right-Angled Triangle Visualization

Instead of using identities, you can find missing ratios by sketching a reference right triangle with the known sides (opposite = 3, hypotenuse = 5). Use the Pythagorean theorem to find the adjacent side ($4$), then apply the correct sign based on the quadrant.