Class 10 Maths Notes Chapter 8 — Introduction to Trigonometry#
Trigonometric ratios#
In $\triangle ABC$, right-angled at $B$, for the acute angle $A$:
$$\sin A = \frac{\text{side opposite to } A}{\text{hypotenuse}} = \frac{BC}{AC}$$$$\cos A = \frac{\text{side adjacent to } A}{\text{hypotenuse}} = \frac{AB}{AC}$$
$$\tan A = \frac{\text{side opposite to } A}{\text{side adjacent to } A} = \frac{BC}{AB}$$
Reciprocal and quotient relations:
$\cosec A = \dfrac{1}{\sin A}$, $\sec A = \dfrac{1}{\cos A}$, $\cot A = \dfrac{1}{\tan A}$
$\tan A = \dfrac{\sin A}{\cos A}$, $\cot A = \dfrac{\cos A}{\sin A}$
- The ratios depend only on the angle, not on the size of the triangle.
- One ratio known, find the rest: if $\sin A = \frac{3}{4}$, take $BC = 3k$ and $AC = 4k$, find $AB$ by Pythagoras, then read off every ratio.
- $\sin A$ and $\cos A$ are never more than 1. $\sec A$ and $\cosec A$ are always 1 or more. (The hypotenuse is the longest side.)
- “$\sin A$” is one symbol. It is not “$\sin$ × $A$”.
Ratios of specific angles#
| $0°$ | $30°$ | $45°$ | $60°$ | $90°$ | |
|---|---|---|---|---|---|
| $\sin$ | $0$ | $\frac{1}{2}$ | $\frac{1}{\sqrt{2}}$ | $\frac{\sqrt{3}}{2}$ | $1$ |
| $\cos$ | $1$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{\sqrt{2}}$ | $\frac{1}{2}$ | $0$ |
| $\tan$ | $0$ | $\frac{1}{\sqrt{3}}$ | $1$ | $\sqrt{3}$ | Not defined |
| $\cosec$ | Not defined | $2$ | $\sqrt{2}$ | $\frac{2}{\sqrt{3}}$ | $1$ |
| $\sec$ | $1$ | $\frac{2}{\sqrt{3}}$ | $\sqrt{2}$ | $2$ | Not defined |
| $\cot$ | Not defined | $\sqrt{3}$ | $1$ | $\frac{1}{\sqrt{3}}$ | $0$ |
To remember the sine row: write $\sqrt{\frac{0}{4}}, \sqrt{\frac{1}{4}}, \sqrt{\frac{2}{4}}, \sqrt{\frac{3}{4}}, \sqrt{\frac{4}{4}}$. The cosine row is the same in reverse order. Then $\tan = \frac{\sin}{\cos}$.
- From 0° to 90°, $\sin A$ increases from 0 to 1 and $\cos A$ decreases from 1 to 0.
- $\tan 90°$, $\sec 90°$, $\cot 0°$ and $\cosec 0°$ are not defined (division by zero).
- If $\tan(A + B) = \sqrt{3}$, then $A + B = 60°$. Match a value to its angle using the table.
Trigonometric identities#
$$\sin^2 A + \cos^2 A = 1 \qquad (0° \leq A \leq 90°)$$$$1 + \tan^2 A = \sec^2 A \qquad (0° \leq A < 90°)$$
$$1 + \cot^2 A = \cosec^2 A \qquad (0° < A \leq 90°)$$
Useful rearrangements:
$\sin^2 A = 1 - \cos^2 A$ and $\cos^2 A = 1 - \sin^2 A$
$\sec^2 A - \tan^2 A = 1$ and $\cosec^2 A - \cot^2 A = 1$
Proving identities:
- Start from the more complicated side (usually the LHS) and work towards the other.
- If stuck, write everything in terms of $\sin$ and $\cos$.
- Look for $(1 - \sin A)(1 + \sin A) = \cos^2 A$ and similar products.
- If the hint says so, simplify LHS and RHS separately to the same expression.
Common mistakes#
- Writing $\sin^2 A$ as $\sin A^2$. $\sin^2 A$ means $(\sin A)^2$.
- Thinking $\sin(A + B) = \sin A + \sin B$. It is false: $\sin 60° \neq \sin 30° + \sin 30°$.
- Mixing up opposite and adjacent sides when the angle changes from $A$ to $C$.
- Forgetting that $\tan 90°$ is not defined, not 0 or ∞.
- Squaring a ratio without squaring the whole fraction: $\left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}$.