Introduction to Trigonometry — chapter notes

In triangle $ABC$, right-angled at $B$, seen from the acute angle $A$: $BC$ is the side opposite $A$, $AB$ the side adjacent to $A$, and $AC$ the hypotenuse.

$\sin A = \frac{BC}{AC}, \quad \cos A = \frac{AB}{AC}, \quad \tan A = \frac{BC}{AB},$ $\operatorname{cosec} A = \frac{1}{\sin A}, \quad \sec A = \frac{1}{\cos A}, \quad \cot A = \frac{1}{\tan A}, \quad \tan A = \frac{\sin A}{\cos A}.$

$\sin A$ is one symbol (not "$\sin \times A$"); $\sin^2 A$ means $(\sin A)^2$. For an acute angle, $0 < \sin A < 1$ and $0 < \cos A < 1$ (the hypotenuse is the longest side), while $\tan A$ can be any positive number.

Two right triangles with the same acute angle $A$ are similar (AA criterion), so corresponding sides are proportional and each ratio has the same value in both. The ratios depend only on the angle.

One ratio given: write two sides as multiples of $k$, find the third by Pythagoras, then read every ratio. Example: $\tan A = \dfrac{5}{12}$ gives sides $5k$, $12k$, $13k$, so $\sin A = \dfrac{5}{13}$, $\cos A = \dfrac{12}{13}$.

Sides from a ratio: a ratio fixes the shape only. $\sin A = \dfrac35$ allows opposite $3$ cm and hypotenuse $5$ cm, or $6$ cm and $10$ cm.

• $45^\circ$: an isosceles right triangle with legs $a$ has hypotenuse $\sqrt2\,a$: $\sin 45^\circ = \cos 45^\circ = \dfrac{1}{\sqrt2}$, $\tan 45^\circ = 1$.
• $30^\circ$ and $60^\circ$: half of an equilateral triangle of side $2a$ has legs $a$ and $\sqrt3\,a$: $\sin 30^\circ = \dfrac12$, $\cos 30^\circ = \dfrac{\sqrt3}{2}$, $\sin 60^\circ = \dfrac{\sqrt3}{2}$, $\cos 60^\circ = \dfrac12$, $\tan 30^\circ = \dfrac{1}{\sqrt3}$, $\tan 60^\circ = \sqrt3$.
• $0^\circ$ and $90^\circ$: $\sin 0^\circ = 0$, $\cos 0^\circ = 1$, $\sin 90^\circ = 1$, $\cos 90^\circ = 0$; $\tan 90^\circ$, $\sec 90^\circ$, $\operatorname{cosec} 0^\circ$, $\cot 0^\circ$ are not defined.

As $A$ increases from $0^\circ$ to $90^\circ$, $\sin A$ increases from $0$ to $1$ and $\cos A$ decreases from $1$ to $0$.

Not in the 2026–27 syllabus: trigonometric ratios of complementary angles.

Evaluating: write each value, square or multiply as the expression says, then simplify. $4\sin^2 30^\circ + 3\tan^2 60^\circ - 2\cos^2 45^\circ = 1 + 9 - 1 = 9$.

Finding angles: read the table backwards. $\sqrt3\tan\theta = 1 \Rightarrow \tan\theta = \dfrac{1}{\sqrt3} \Rightarrow \theta = 30^\circ$. Two equations such as $\sin(A - B) = \dfrac12$, $\cos(A + B) = \dfrac12$ give $A - B = 30^\circ$, $A + B = 60^\circ$, so $A = 45^\circ$, $B = 15^\circ$.

Sides of a triangle: choose the ratio that links the known side and the wanted side. With $QR = 9$ cm opposite $\angle P = 60^\circ$: $PQ = \dfrac{9}{\tan 60^\circ} = 3\sqrt3$ cm and $PR = \dfrac{9}{\sin 60^\circ} = 6\sqrt3$ cm.

Sines do not add: $\sin 30^\circ + \sin 60^\circ \ne \sin 90^\circ$.

In triangle $ABC$, right-angled at $B$, $AB^2 + BC^2 = AC^2$. Dividing by $AC^2$, $AB^2$ and $BC^2$ in turn:

$\sin^2 A + \cos^2 A = 1 \quad (0^\circ \le A \le 90^\circ)$ $1 + \tan^2 A = \sec^2 A \quad (0^\circ \le A < 90^\circ)$ $\cot^2 A + 1 = \operatorname{cosec}^2 A \quad (0^\circ < A \le 90^\circ)$

An identity is true for every angle for which both sides are defined. With them, one ratio gives all the others without a triangle: $\sec\theta = \dfrac{25}{7}$ gives $\tan\theta = \dfrac{24}{7}$, $\cos\theta = \dfrac{7}{25}$, $\sin\theta = \dfrac{24}{25}$.

Work on one side (usually the more complicated) until it becomes the other; never move terms across "=".

1. Replace a bracket by an identity when one matches: $1 + \tan^2 A = \sec^2 A$, $1 - \sin^2 A = \cos^2 A$.
2. Otherwise write everything in $\sin$ and $\cos$, then combine fractions.
3. Factorise: $\sin^4 A - \cos^4 A = (\sin^2 A - \cos^2 A)(\sin^2 A + \cos^2 A)$.
4. Multiply by a conjugate: $\dfrac{1 + \sin A}{\cos A} \times \dfrac{1 - \sin A}{1 - \sin A} = \dfrac{\cos A}{1 - \sin A}$.

Conditions: a relation like $5\sin\theta = 12\cos\theta$ gives $\tan\theta = \dfrac{12}{5}$, then the identities give the rest. Squaring sums: $(\sin\theta + \cos\theta)^2 = 1 + 2\sin\theta\cos\theta$, so $\sin\theta + \cos\theta \le \sqrt2$.

Test with $30^\circ$, $45^\circ$, $60^\circ$ to catch errors — but numbers never replace the proof.