ICSE Class 9 Maths Chapter 23 Trigonometrical Ratios of Standard Angles

Trigonometrical Ratios Of Standard Angles

Including Evaluation of an Expression Involving Trigonometric Ratios

23.1 Trigonometrical Ratios Of Angles 30° And 60°

Let ABC be an equilateral triangle with side '2a' and AD is perpendicular to BC.

Clearly, BD = BC/2 = a

In triangle ABD, AD² = AB² - BD² (Using Pythagoras Theorem)

= (2a)² - a² = 3a²

Therefore, AD = \(\sqrt{3} \cdot a\)

Since, each angle of an equilateral triangle is 60°

Therefore, angle B = 60° and

angle BAD = 180° - (60° + 90°) = 30°

Therefore, In triangle ABD,

\(\sin 60° = \frac{\text{perp.}}{\text{hyp.}} = \frac{AD}{AB} = \frac{\sqrt{3} \cdot a}{2a} = \frac{\sqrt{3}}{2}\) ; \(\sin 30° = \frac{\text{perp.}}{\text{hyp.}} = \frac{BD}{AB} = \frac{a}{2a} = \frac{1}{2}\)

\(\cos 60° = \frac{\text{base}}{\text{hyp.}} = \frac{BD}{AB} = \frac{a}{2a} = \frac{1}{2}\) ; \(\cos 30° = \frac{\text{base}}{\text{hyp.}} = \frac{AD}{AB} = \frac{\sqrt{3} \cdot a}{2a} = \frac{\sqrt{3}}{2}\)

\(\tan 60° = \frac{\text{perp.}}{\text{base}} = \frac{AD}{BD} = \frac{\sqrt{3} \cdot a}{a} = \sqrt{3}\) ; \(\tan 30° = \frac{\text{perp.}}{\text{base}} = \frac{BD}{AD} = \frac{a}{\sqrt{3} \cdot a} = \frac{1}{\sqrt{3}}\)

and so on.

23.2 Trigonometrical Ratios Of Angle 45°

The adjoining figure shows a right-angled isosceles triangle in which angle B = 90° and AB = BC = a.

Clearly, angle A = 45°

and, AC = \(\sqrt{2} \cdot a\) (Since, AC² = AB² + BC²)

Therefore, \(\sin 45° = \frac{\text{perp.}}{\text{hyp.}} = \frac{BC}{AC} = \frac{a}{\sqrt{2} \cdot a} = \frac{1}{\sqrt{2}}\) ;

\(\cos 45° = \frac{\text{base}}{\text{hyp.}} = \frac{AB}{AC} = \frac{a}{\sqrt{2} \cdot a} = \frac{1}{\sqrt{2}}\)

\(\tan 45° = \frac{\text{perp.}}{\text{base}} = \frac{BC}{AB} = \frac{a}{a} = 1\) and so on.

Also, remember that:

\(\sin 0° = 0\); \(\cos 0° = 1\)

\(\sin 90° = 1\); \(\cos 90° = 0\)

Teacher's Note

Understanding standard angles helps engineers and architects quickly calculate angles without a calculator, similar to how a carpenter uses common angles like 45° and 90° daily.

Trigonometric Ratios Table For Standard Angles

Therefore, for standard angles of 0°, 30°, 45°, 60° and 90°, we have:

Important Observations

1. It can be seen from the table, given above, that as the angle increases from 0° to 90°:

(i) value of sin increases from 0 to 1

(ii) value of cos decreases from 1 to 0

(iii) value of tan increases from 0 to \(\infty\) and so on.

2. If x = y = 45° or, x + y = 90° then, sin x = cos y; tan x = cot y and sec x = cosec y.

For example: \(\sin 45° = \cos 45° = \frac{1}{\sqrt{2}}\) [x = y = 45°]

and also \(\sin 30° = \cos 60° = \frac{1}{2}\) [x + y = 90°]

3. Whatever be the measure of angle A;

(i) \(\sin^2 A + \cos^2 A = 1\)

(ii) \(\sec^2 A - \tan^2 A = 1\)

(iii) \(\text{cosec}^2 A - \cot^2 A = 1\)

For example: If A = 30°; \(\sin^2 A + \cos^2 A = \sin^2 30° + \cos^2 30°\)

\(= \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2\)

\(= \frac{1}{4} + \frac{3}{4} = 1\)

Similarly, \(\sec^2 A - \tan^2 A = \sec^2 30° - \tan^2 30°\)

\(= \left(\frac{2}{\sqrt{3}}\right)^2 - \left(\frac{1}{\sqrt{3}}\right)^2\)

\(= \frac{4}{3} - \frac{1}{3} = 1\)

Teacher's Note

These fundamental trigonometric identities are as universal as the Pythagorean theorem - they work for any angle, just like how gravity affects all objects equally regardless of their size.

Important Identities

1. \(\sin (A + B) \neq \sin A + \sin B\)

2. \(\cos (A + B) \neq \cos A + \cos B\)

3. \(\tan (A + B) \neq \tan A + \tan B\)

4. \(\sin (A - B) \neq \sin A - \sin B\) and so on.

Example Problems

Problem 1: Evaluate

(i) \(\sin^2 30° - 2 \cos^3 60° + 3 \tan^4 45°\)

(ii) \((\cos 0° + \sin 45° + \sin 30°)(\sin 90° - \cos 45° + \cos 60°)\)

Solution

(i) \(\sin^2 30° - 2 \cos^3 60° + 3 \tan^4 45°\)

\(= \left(\frac{1}{2}\right)^2 - 2\left(\frac{1}{2}\right)^3 + 3(1)^4\)

\(= \frac{1}{4} - 2 \times \frac{1}{8} + 3 \times 1 = \frac{1}{4} - \frac{1}{4} + 3 = 3\)

(ii) \((\cos 0° + \sin 45° + \sin 30°)(\sin 90° - \cos 45° + \cos 60°)\)

\(= \left(1 + \frac{1}{\sqrt{2}} + \frac{1}{2}\right)\left(1 - \frac{1}{\sqrt{2}} + \frac{1}{2}\right)\)

\(= \left(\frac{3}{2} + \frac{1}{\sqrt{2}}\right)\left(\frac{3}{2} - \frac{1}{\sqrt{2}}\right)\)

\(= \left(\frac{3}{2}\right)^2 - \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{9}{4} - \frac{1}{2} = 1\frac{3}{4}\)

Problem 2: Find the value of

\[\frac{\sin 30° - \sin 90° + 2 \cos 0°}{\tan 30° \times \tan 60°}\]

Solution

\[\frac{\sin 30° - \sin 90° + 2 \cos 0°}{\tan 30° \times \tan 60°} = \frac{\frac{1}{2} - 1 + 2 \times 1}{\frac{1}{\sqrt{3}} \times \sqrt{3}}\]

\[= \frac{\frac{1}{2} - 1 + 2}{1}\]

\[= \frac{3}{2} = 1\frac{1}{2}\]

Problem 3: If A = 60°, verify that

(i) \(\sin^2 A + \cos^2 A = 1\)

(ii) \(\sec^2 A - \tan^2 A = 1\)

Solution

(i) \(\sin^2 A + \cos^2 A = \sin^2 60° + \cos^2 60° = \left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{3}{4} + \frac{1}{4} = 1\)

(ii) \(\sec^2 A - \tan^2 A = \sec^2 60° - \tan^2 60° = (2)^2 - (\sqrt{3})^2 = 4 - 3 = 1\)

Problem 4: If x = 15°, evaluate: 8 sin 2x cos 4x sin 6x

Solution

8 sin 2x cos 4x sin 6x = 8 sin(2 × 15°) cos(4 × 15°) sin(6 × 15°)

= 8 sin 30° cos 60° sin 90°

\(= 8 \times \frac{1}{2} \times \frac{1}{2} \times 1 = 2\)

Teacher's Note

These evaluation problems train the mind to recognize patterns and substitute known values systematically, much like a musician sight-reading music by recognizing familiar note patterns.

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