Class 10 NCERT Maths Chapter 8: Introduction to Trigonometry
Trigonometry studies the relationship between the angles and sides of a right-angled triangle. It’s widely used in navigation, architecture, astronomy, and even in GPS technology.
Trigonometric Ratios
In a right-angled triangle:
- Sine: \(\sin \theta = \frac{\text{Perpendicular (p)}}{\text{Hypotenuse (h)}}\)
- Cosine: \(\cos \theta = \frac{\text{Base (b)}}{\text{Hypotenuse (h)}}\)
- Tangent: \(\tan \theta = \frac{\text{Perpendicular (p)}}{\text{Base (b)}}\)
- Cosecant: \(\csc \theta = \frac{\text{Hypotenuse (h)}}{\text{Perpendicular (p)}}\)
- Secant: \(\sec \theta = \frac{\text{Hypotenuse (h)}}{\text{Base (b)}}\)
- Cotangent: \(\cot \theta = \frac{\text{Base (b)}}{\text{Perpendicular (p)}}\)
Standard Trigonometric Values
| \(\theta\) | \(\sin\theta\) | \(\cos\theta\) | \(\tan\theta\) | \(\csc\theta\) | \(\sec\theta\) | \(\cot\theta\) |
|---|---|---|---|---|---|---|
| 0° | 0 | 1 | 0 | Not defined | 1 | Not defined |
| 30° | \(\frac{1}{2}\) | \(\frac{\sqrt{3}}{2}\) | \(\frac{1}{\sqrt{3}}\) | 2 | \(\frac{2}{\sqrt{3}}\) | \(\sqrt{3}\) |
| 45° | \(\frac{1}{\sqrt{2}}\) | \(\frac{1}{\sqrt{2}}\) | 1 | \(\sqrt{2}\) | \(\sqrt{2}\) | 1 |
| 60° | \(\frac{\sqrt{3}}{2}\) | \(\frac{1}{2}\) | \(\sqrt{3}\) | \(\frac{2}{\sqrt{3}}\) | 2 | \(\frac{1}{\sqrt{3}}\) |
| 90° | 1 | 0 | Not defined | 1 | Not defined | 0 |
Trigonometric Identities
For \(0^\circ < \theta < 90^\circ\):
\[ \sin^2\theta + \cos^2\theta = 1 \] \[ \sec^2\theta - \tan^2\theta = 1 \] \[ \csc^2\theta - \cot^2\theta = 1 \]Quick Example
Example: If \(\sin\theta = \frac{p}{h} = \frac{3}{5}\), find \(\cos\theta\).
Using \(\sin^2\theta + \cos^2\theta = 1\): \[ \left(\frac{3}{5}\right)^2 + \cos^2\theta = 1 \] \[ \frac{9}{25} + \cos^2\theta = 1 \] \[ \cos^2\theta = \frac{16}{25} \] \(\Rightarrow \cos\theta = \frac{4}{5} = \frac{b}{h}\)
📝 Memory Tip
Remember the short forms:
- \(\sin\theta = p/h\)
- \(\cos\theta = b/h\)
- \(\tan\theta = p/b\)
- \(\csc\theta = h/p\)
- \(\sec\th