Multiple Angles

The multiple angles topic comes under the trigonometric functions. It is not possible to find the values of multiple angles directly. We can calculate the values of multiple angles by expressing each trigonometric function in its expanded form. Learning the multiple angle formulas helps students to save time while solving problems. In this article, we discuss the formula for multiple angles in trigonometry.

Let A be a given angle, then 2A, 3A, 4A, etc., are called multiple angles. The double and triple angles formula are used under the multiple angle formulas. The common functions in trigonometry that are used for the multiple angle formula are sine, cosine and tangent.

Multiple Angle Formulas

1) The sin formula for multiple angles is given by

Where n = 1,2,3..

2) The cosine formula for multiple angles is given by

Where n = 1,2,3..

3) The tangent formula for multiple angles is given by

tan nθ = sin nθ/cos nθ

The important trigonometric ratios of multiple angle formulae are as follows:

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

(b) cos 2A = cos²A – sin²A

(c) sin 2A = 2 tan A/(1+tan²A)

(d) cos 2A = 2 cos²A – 1

(e) cos 2A = 1 – 2 sin²A

(f) 2 cos² A = 1 + cos 2A

(g) 2 sin² A = 1 – cos 2A

(h) tan² A = (1 – cos 2A)/(1 + cos 2A)

(i) cos 2A = (1 – tan² A)/(1 + tan² A)

(j) tan 2A = 2 tan A/(1 – tan² A)

(k) sin 3A = 3 sin A – 4 sin³A

(l) cos 3A = 4 cos³A – 3 cos A

(m) tan 3A = (3 tan A – tan³A)/(1 – 3 tan²A)

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

(i) 2 sin^-1x = sin^-1(2x√(1-x²))

(ii) 2 cos^-1x = cos^-1(2x²-1)

(iii) 2 tan^-1x = tan^-1(2x/(1-x²)) = sin^-1(2x/(1+x²) = cos^-1(1-x²)/(1+x²)

(iv) 3 sin^-1x = sin^-1(3x-4x³)

(v) 3 cos^-1x = cos^-1(4x³-3x)

(vi) 3 tan^-1x = tan^-1[(3x-x³)/(1-3x²)]

Also Read

Trigonometric Equations and Its Solutions

JEE Inverse Trig Functions Previous Year Questions with Solutions

Solved Examples Using Multiple Angle Formulas

Example 1:

Show that (3 sin A-4 sin³A)/(4 cos³ A – 3 cos A) = tan 3A by using the multiple angle formula.

Solution:

We have sin 3A = 3 sin A-4 sin³A

cos 3A = 4 cos³ A – 3 cos A

(3 sin A-4 sin³A)/(4 cos³ A – 3 cos A) = sin 3A/cos 3A

= tan 3A

Hence proved.

Example 2:

Calculate sin 2α, cos 2α, and tan 2α if sin α = 1/2 and the angle α lies in the 1st quadrant.

Solution:

Given sin α = 1/2

So cos α = √(1 – sin²α)

= √(1 – 1/4)

= √3/2

tan α = sin α/cos α

= 1/√3

sin 2α = 2 sin α cos α

= 2.(1/2).√3/2

= √3/2

cos 2α = cos²α – sin²α

= (¾) – (¼)

= 1/2

tan 2α = sin 2α/cos 2α

= √3

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