CBSE Class 10 Maths Notes: Introduction to Trigonometry

  Trigonometric Ratios: Definitions

Trigonometry is the study of the relationships between the sides and angles of triangles. In this chapter, we’ll focus on right-angled triangles. Let’s define the core trigonometric ratios for an acute angle (an angle less than 90 degrees) within a right triangle. Consider a right triangle ABC, right-angled at B. For an acute angle, let’s say angle $A$ (denoted by $\theta$):

• Sine (sin): $\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC}$
• Cosine (cos): $\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC}$
• Tangent (tan): $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BC}{AB}$
• Cosecant (csc or cosec): $\csc(\theta) = \frac{1}{\sin(\theta)} = \frac{\text{Hypotenuse}}{\text{Opposite}} = \frac{AC}{BC}$
• Secant (sec): $\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{\text{Hypotenuse}}{\text{Adjacent}} = \frac{AC}{AB}$
• Cotangent (cot): $\cot(\theta) = \frac{1}{\tan(\theta)} = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{AB}{BC}$

  Well-Definedness of Trigonometric Ratios

The key point is: these ratios are well-defined. This means the ratios (sin, cos, tan, etc.) depend only on the angle $\theta$ and not on the size of the right triangle.

Why? Consider two similar right-angled triangles with the same acute angle $\theta$. Since the triangles are similar, their corresponding sides are proportional. If we double the sides of the triangle, both the adjacent, opposite, and hypotenuse lengths double, but the ratio (e.g., $\frac{\text{Opposite}}{\text{Hypotenuse}}$) remains unchanged. This is because the ratios only depend on the angle. This is based on the idea of similar triangles, and the proportionality of their sides.

  Trigonometric Ratios for Standard Angles

It is essential to memorize the values of trigonometric ratios for the following standard angles: $0^\circ, 30^\circ, 45^\circ, 60^\circ,$ and $90^\circ$. While you can derive these values (e.g., using an equilateral triangle for $60^\circ$ and a square for $45^\circ$), memorization is highly recommended.

Here’s a table for 30°, 45°, and 60° (values for 0° and 90° are also important):

Angle ($\theta$)	sin($\theta$)	cos($\theta$)	tan($\theta$)	csc($\theta$)	sec($\theta$)	cot($\theta$)
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	$\frac{1}{\sqrt{3}}$

  Basic Trigonometric Relationships

These relationships are crucial for simplifying expressions and solving equations.

• Reciprocal Relations:
  • $\csc(\theta) = \frac{1}{\sin(\theta)}$
  • $\sec(\theta) = \frac{1}{\cos(\theta)}$
  • $\cot(\theta) = \frac{1}{\tan(\theta)}$
• Quotient Relations:
  • $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$
  • $\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$

  The Fundamental Trigonometric Identity

This is the most important identity in trigonometry.

Identity: $\sin^2(A) + \cos^2(A) = 1$

Proof: Consider a right-angled triangle ABC, right-angled at B. Let $\theta = A$.

• By definition, $\sin(A) = \frac{BC}{AC}$ and $\cos(A) = \frac{AB}{AC}$.
• Therefore, $\sin^2(A) = \frac{BC^2}{AC^2}$ and $\cos^2(A) = \frac{AB^2}{AC^2}$.
• Adding these: $\sin^2(A) + \cos^2(A) = \frac{BC^2}{AC^2} + \frac{AB^2}{AC^2} = \frac{BC^2 + AB^2}{AC^2}$.
• By the Pythagorean theorem, in a right-angled triangle: $AB^2 + BC^2 = AC^2$
• So, $\sin^2(A) + \cos^2(A) = \frac{AC^2}{AC^2} = 1$.

  Derived Identities and Manipulations

The fundamental identity can be rearranged to derive other useful identities and to simplify expressions:

• $\sin^2(A) = 1 – \cos^2(A) \implies \sin(A) = \pm\sqrt{1 – \cos^2(A)}$
• $\cos^2(A) = 1 – \sin^2(A) \implies \cos(A) = \pm\sqrt{1 – \sin^2(A)}$

Example: If $\sin(\theta) = \frac{3}{5}$, find $\cos(\theta)$.

Using the identity: $\cos^2(\theta) = 1 – \sin^2(\theta) = 1 – (\frac{3}{5})^2 = 1 – \frac{9}{25} = \frac{16}{25}$. Therefore, $\cos(\theta) = \pm\sqrt{\frac{16}{25}} = \pm\frac{4}{5}$. The sign will depend on the quadrant in which the angle $\theta$ lies. For acute angles, we typically consider the positive value, $\frac{4}{5}$.