Trigonometric Ratios: Practice Problems and Methods

Definitions and special angles

In right triangle ABC with $\angle C = 90^\circ$: $\sin A = \dfrac{BC}{AB}$, $\cos A = \dfrac{AC}{AB}$, $\tan A = \dfrac{BC}{AC}$.

The triangles with side ratios $1 : 2 : \sqrt{3}$ and $1 : 1 : \sqrt{2}$ give $\sin 30^\circ = \frac{1}{2}$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$, $\tan 45^\circ = 1$, and so on.

For the point $P(x,\ y)$ at angle $\theta$ on the unit semicircle, $\sin\theta = y$, $\cos\theta = x$ and $\tan\theta = \frac{y}{x}$. This defines the ratios for obtuse angles too.

Identities

If $\sin\theta = \frac{3}{5}$ and $\theta$ is obtuse, then $\cos^2\theta = 1 - \frac{9}{25} = \frac{16}{25}$, and since $\cos\theta < 0$ for obtuse angles, $\cos\theta = -\frac{4}{5}$. The range of $\theta$ decides the sign.

Trigonometric equations and inequalities

To solve $\sin\theta = \frac{1}{2}$ for $0^\circ \leqq \theta \leqq 180^\circ$, find the points on the unit circle with $y$-coordinate $\frac{1}{2}$: $\theta = 30^\circ,\ 150^\circ$. For $\sin\theta > \frac{1}{2}$, take the part above that height: $30^\circ < \theta < 150^\circ$.