Multiple Angles in Trigonometry - Formulas & Solved Examples | Testbook.com

Last Updated on Jul 31, 2023
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The concept of multiple angles is a significant part of trigonometric functions. These angles are not directly calculable. Instead, we need to express each trigonometric function in its expanded form to find the values of multiple angles. Mastery of the multiple angle formulas can greatly speed up problem-solving in trigonometry. This article will help you understand the formulas for multiple angles in trigonometry in a more detailed manner.

Let's consider an angle A. The multiples of this angle, such as 2A, 3A, 4A, etc., are referred to as multiple angles. The double and triple angle formulas are part of the broader multiple angle formulas. The most commonly used functions in trigonometry for the multiple angle formula are sine, cosine, and tangent.

Formulas for Multiple Angles

Here are the main formulas for multiple angles:

1) The sin formula for multiple angles is given by

Where n can be any integer such as 1,2,3..

2) The cosine formula for multiple angles is given by

Where n can be any integer such as 1,2,3..

3) The tangent formula for multiple angles is given by

tan nθ = sin nθ/cos nθ

Here are some important trigonometric ratios of multiple angle formulae:

(a) sin 2A = 2 sin A cos A

(b) cos 2A = cos 2 A – sin 2 A

(c) sin 2A = 2 tan A/(1+tan 2 A)

(d) cos 2A = 2 cos 2 A – 1

(e) cos 2A = 1 – 2 sin 2 A

(f) 2 cos 2 A = 1 + cos 2A

(g) 2 sin 2 A = 1 – cos 2A

(h) tan 2 A = (1 – cos 2A)/(1 + cos 2A)

(i) cos 2A = (1 – tan 2 A)/(1 + tan 2 A)

(j) tan 2A = 2 tan A/(1 – tan 2 A)

(k) sin 3A = 3 sin A – 4 sin 3 A

(l) cos 3A = 4 cos 3 A – 3 cos A

(m) tan 3A = (3 tan A – tan 3 A)/(1 – 3 tan 2 A)

The formulae of multiple angles for different inverse trigonometric functions are as follows:

(i) 2 sin -1 x = sin -1 (2x√(1-x 2 ))

(ii) 2 cos -1 x = cos -1 (2x 2 -1)

(iii) 2 tan -1 x = tan -1 (2x/(1-x 2 )) = sin -1 (2x/(1+x 2 ) = cos -1 (1-x 2 )/(1+x 2 )

(iv) 3 sin -1 x = sin -1 (3x-4x 3 )

(v) 3 cos -1 x = cos -1 (4x 3 -3x)

(vi) 3 tan -1 x = tan -1 [(3x-x 3 )/(1-3x 2 )]

Further Reading

Solutions to Trigonometric Equations

Past Year Questions on Inverse Trig Functions with Solutions

Applications of Multiple Angle Formulas: Solved Examples

Example 1:

Use the multiple angle formula to prove that (3 sin A-4 sin 3 A)/(4 cos 3 A – 3 cos A) = tan 3A.

Solution:

The given expression can be rewritten using the identities sin 3A = 3 sin A-4 sin 3 A and cos 3A = 4 cos 3 A – 3 cos A as follows:

(3 sin A-4 sin 3 A)/(4 cos 3 A – 3 cos A) = sin 3A/cos 3A

= tan 3A

This confirms the given identity.

Example 2:

If sin α = 1/2 and the angle α is in the 1st quadrant, find sin 2α, cos 2α, and tan 2α.

Solution:

Given that sin α = 1/2, we can find cos α using the Pythagorean identity:

cos α = √(1 – sin 2 α) = √(1 – 1/4) = √3/2

Then, tan α = sin α/cos α = 1/√3

Using the double angle formulas, we can find:

sin 2α = 2 sin α cos α = 2.(1/2).√3/2 = √3/2

cos 2α = cos 2 α – sin 2 α = (¾) – (¼) = 1/2

tan 2α = sin 2α/cos 2α = √3

More Articles for Maths

Frequently Asked Questions

sin 2A = 2 sin A cos A, cos 2A = cos^2 A – sin^2 A, tan 2A = (2 tan A)/(1-tan^2 A)

sin 3A = 3 sin A – 4 sin^3 A, cos 3A = 4 cos^3 A – 3 cos A, tan 3A = (3 tan A – tan^3 A)/(1 – 3 tan^2 A)

2 sin^-1 x = sin^-1 (2x√(1-x^2)), 2 cos^-1 x = cos^-1 (2x^2 -1), 2 tan^-1 x = tan^-1 (2x/(1-x^2)), 3 sin^-1 x = sin^-1 (3x-4x^3), 3 cos^-1 x = cos^-1 (4x^3 -3x), 3 tan^-1 x = tan^-1 [(3x-x^3 )/(1-3x^2 )]


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