Multiple Angle Formulas: Sine, Cosine & Tangent

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Multiple angles, in trigonometry, are commonly found in trigonometric functions. It is not possible to directly determine the values of many angles, but they can be determined by expressing each trigonometric function in its extended form.

The multiple-angle formulas include the double and triple-angle formulas. The common functions employed in the multiple-angle formula include sine, tangent, and cosine.
In mathematics, multiple angles are the types of angles that are a multiple of a given angle.
For example, the sine and cosine of multiple angles can be represented in terms of the sine and cosine of the given angle using trigonometric identities.
Multiple angle formulas can be used to simplify trigonometric expressions and to find exact solutions to trigonometric equations.

Table of Content

What are Multiple Angles?
Multiple Angles Formula
Multiple Angle Formula for Inverse Trigonometric Functions
Things to Remember
Previous Year Questions
Sample Questions

Key Terms: Multiple Angles, Trigonometric Functions, Double Angle, Triple Angle, Sine, Tangent, Trigonometric Equations, Binomial Theorem, Euler’s Formula

What are Multiple Angles?

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Multiple angles can be seen in Trigonometric Functions. The values of the multiple angles cannot be found directly, yet their values can be evaluated by expressing every trigonometric function in its expanded form.

The trigonometric function of multiple angles is known as the multiple-angle formula.
The double and triple angles formula are often applicable under the Multiple-Angle formulas.
It's important to note that while multiple-angle formulas help simplify trigonometric expressions, they should be used with care as they can sometimes lead to numerical instability or loss of precision.
It's usually better to use these formulas in conjunction with other techniques, such as factoring or substitution, to simplify trigonometric expressions.
Sine, Tangent and Cosine are the common functions used for the multiple-angle formula.

Using Euler's formula and the Binomial Theorem, one can calculate multiple angles of the type sin nx, cos nx, and tan nx, which are all represented in terms of sin x. The double and triple-angle formulas are examples of multiple-angle formulas. The multiple-angle formula is another name for the trigonometric function of multiple angles.

Multiple Angles Formulas

Multiple Angles Formulas

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Multiple Angles Formula

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Multiple angle formulas are used in trigonometry in order to express trigonometric functions of angles which are multiples of a given angle in terms of trigonometric functions of the given angle. With the help of these formulas, trigonometric expressions can be simplified and trigonometric equations can be solved.

Sine Formula

For multiple angles, the sin formula is as follows:

\(sin \text{ n}\theta = \displaystyle\sum_{k=0}^{n} cos^{k} \theta \text{ sin}^{n-k}\theta \text{ sin} [\frac{1}{2}(n-k)\pi]\)

Where n = 1, 2, 3...

The following are some basic formulas:

Sin 2θ = 2 × Cosθ.Sinθ
Sin 3θ = 3 Sinθ – 4 Sin3θ

Cosine Formula

For multiple angle, the cosine formula is:

\(cos \text{ n}\theta = \displaystyle\sum_{k=0}^{n} cos^{k} \theta \text{ sin}^{n-k}\theta \text{ cos} [\frac{1}{2}(n-k)\pi]\)

Where n = 1, 2, 3…

The following are basic formulas:

Cos 2θ = Cos 2θ – Sin 2θ
Cos 3θ = 4Cos 3θ – 3Cos θ

Tangent Formula

For multiple angle, the tangent formula is:

\(Tan \text{ n}\theta = \frac{Sin\text{ n}\theta}{Cos\text{ n}\theta}\)

Where n = 1, 2, 3..

Its basic formula can be represented as:

→ tan nθ = sin nθ/cos nθ

Trigonometric Ratios for Multiple Angles

The following are the important trigonometric ratios for multiple angle equations.

sin 2A = 2 sin A cos A
cos 2A = cos2A – sin2A
sin 2A = 2 tan A/(1 + tan2A)
cos 2A = 2 cos2A – 1
cos 2A = 1 – 2 sin2A
2 cos2 A = 1 + cos 2A
2 sin2 A = 1 – cos 2A
tan2 A = (1 – cos 2A)/(1 + cos 2A)
cos 2A = (1 – tan2 A)/(1 + tan2 A)
tan 2A = 2 tan A/(1 – tan2 A)
sin 3A = 3 sin A – 4 sin3A
cos 3A = 4 cos3A – 3 cos A
tan 3A = (3 tan A – tan3A)/(1 – 3 tan2A)

Trigonometric Functions Detailed Video Explanation

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Multiple Angle Formula for Inverse Trigonometric Functions

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The following are multiple angle formulae for various inverse trigonometric functions.

2 sin-1x = sin-1(2x√(1-x2))
2 cos-1x = cos-1(2x2-1)
2 tan-1x = tan-1(2x/(1-x2)) = sin-1(2x/(1+x2) = cos-1(1-x2)/(1+x2)
3 sin-1x = sin-1(3x-4x3)
3 cos-1x = cos-1(4x3-3x)
3 tan-1x = tan-1[(3x-x3)/(1-3x2)]

Things to Remember

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Values of multiple angles can be determined by expressing each trigonometric function in its extended form.
The multiple-angle formulas include the double and triple-angle formulas.
The common functions used in multiple angles, in trigonometry, include sine, tangent, and cosine.
The multiple-angle formula is another name for the trigonometric function of multiple angles.
The sin formula for multiple angles is: \(sin \text{ n}\theta = \displaystyle\sum_{k=0}^{n} cos^{k} \theta \text{ sin}^{n-k}\theta \text{ sin} [\frac{1}{2}(n-k)\pi]\).
The cos formula For multiple angles is: \(cos \text{ n}\theta = \displaystyle\sum_{k=0}^{n} cos^{k} \theta \text{ sin}^{n-k}\theta \text{ cos} [\frac{1}{2}(n-k)\pi]\).
The tan formula For multiple angles is: \(Tan \text{ n}\theta = \frac{Sin\text{ n}\theta}{Cos\text{ n}\theta}\).

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Sample Questions

Ques: Prove that cos 5x = 16 cos⁵x – 20 cos³x + 5 cos x: (4 marks)

Ans: L.H.S. = cos 5x

= cos (2x + 3x)

= cos 2x·cos 3x - sin 2x·sin 3x

= (2 cos² x - 1) (4 cos³ x - 3 cos x) - 2 sin x cos x (3 sin x - 4 sin³ x)

= 8 cos⁵ x - 10 cos³ x + 3 cos x - 6 cos x sin² x + 8 cos x sin⁴ x

= 8 cos⁵ x - 10 cos³ x + 3 cos x - 6 cos x (1 - cos² x) + 8 cos x (1 - cos² x)²

= 8 cos⁵ x - 10 cos³ x + 3 cos x - 6 cos x + 6 cos³x + 8 cos x - 16 cos³ x + 8 cos⁵ x

= 16 cos⁵ x - 20 cos³ x + 5 cos x

Ques: Prove that \((\frac{sin\,2x + sin\,x}{cos\,2x+1+cos\,x}=tan\,x)\). (3 marks)

Ans: Given: Using the multiple angle formulas,

Sin2θ =2×Cosθ.Sinθ and, Cos2θ = Cos2θ – Sin2θ

\(\frac{sin\,2x + sin\,x}{cos\,2x+1+cos\,x}=tan\,x\)

Putting the values in L.H.S

\(\frac{2\,sin\,x\;cos\,x+sin\,x}{cos\,x+2cos^{2}\,x}\)

\(=\frac{sin\,x(1+2cos\,x)}{cos\,x(2\,cos\,x+1)}\)

R.H.S = tanx

Hence proved, \(\frac{sin\,x+sin\,2x}{1+cos\,x+cos\,2x}=tan\,x\)

Ques: Using the multiple angle formula, show that (3 sin A-4 sin3A)/(4 cos3 A – 3 cos A) = tan 3A. (2 marks)

Ans: We can deduce that sin 3A = 3 sin A-4 sin3A.

cos 3A = 4 cos

3 A – 3 cos A

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

tan 3A. proved.

Ques: If sin = 1/2 and the angle is in the first quadrant, calculate sin 2, cos 2, and tan 2. (4 marks)

Ans: 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

Ques: Prove that \((\frac{3 \sin\theta -4 sin^{3}\theta}{4 cos^{3}\, \theta – 3 cos\, \theta}=tan\;3\theta) \) by using multiple angle formulas. (3 marks)

Ans: Using the multiple-angle formulas,

Sin3θ = 3Sinθ - 4Sin3θ, and Cos3θ =4Cos3θ – 3Cosθ

Putting the values in L.H.S,

\(=\frac{3Sin\theta - 4Sin^3\theta}{4Cos^3\theta – 3Cos\theta}\)

\(=\frac{Sin3\theta}{Cos3\theta}\)

R.H.S = tan3θ

Hence proved, \(\frac{3 \sin\theta -4 sin^{3}\theta}{4 cos^{3}\, \theta – 3 cos\, \theta}=tan\;3\theta\)

Ques: Express tan(3α)tan(3α) in terms of “tanα tanα”. (1 mark)

Ans: tan(3α)=tan(2α+α)

=tan(2α)+tanα1−tan(2α)tanα

=2tanα1−tan2α+tanα1−2tanα1−tan2α⋅tanα

=2tanα+tanα(1−tan2α)1−tan2α−2tanα⋅tanα

=3tanα−tan3α1−3tan2α

Ques: What are the formulas for double angles? (3 marks)

Ans: The trigonometric ratios of double angles (2) are written using double-angle formulas in terms of trigonometric ratios of single angles.

The following are the double-angle formulas for sin, cos, and tan:sin

2A = 2 sin A cos A or,

sin 2A = (2 tan A) / (1 + tan2A)

cos 2A = cos2A - sin2A or,

cos 2A = 2cos2A - 1 or,

cos 2A = 1 - 2sin2A or,

cos 2A = (1 - tan2A) / (1 + tan2A)

tan 2A = (2 tan A)/(1 - tan2A)

Ques: Find sin2θ if sin θ = ?. (3 marks)

Ans: We are aware of this.

sin2θ = 2 sin θ cos θ

We need to figure out what cos θ. is.

Let's utilise the sin cos formula cos2θ + sin2θ = 1.

We get cos2θ = 1 - sin2θ

= 1-(9/25)

cos2θ = 16/25

cos θ = 4/5

sin2θ = 2sinθcos θ

= 2 × (3/5) × (4/5) = 24/25

Ques. Prove the following: i) (sin⁴θ – cos⁴θ +1) cosec²θ = 2
ii) (√3 + 1) (3 – cot 30°) = tan³60° – 2 sin 60°. (5 marks)

Ans. i) As per the L.H.S. = (sin⁴θ – cos⁴θ +1) cosec²θ

Thus,

= [(sin²θ – cos²θ) (sin²θ + cos²θ) + 1] cosec²θ

By applying the identity sin²A + cos²A = 1, we get

= (sin²θ – cos²θ + 1) cosec²θ

= [sin²θ – (1 – sin²θ) + 1] cosec²θ

= 2 sin²θ cosec²θ

= 2 sin²θ (1/sin²θ)

= 2

= RHS

ii) As per the LHS,

LHS = (√3 + 1)(3 – cot 30°)

= (√3 + 1)(3 – √3)

= 3√3 – √3.√3 + 3 – √3

= 2√3 – 3 + 3

= 2√3

Then, we can see that:

RHS = tan³60° – 2 sin 60°

= (√3)³ – 2(√3/2)

= 3√3 – √3

= 2√3

Thus, (√3 + 1) (3 – cot 30°) = tan³60° – 2 sin 60°.

Hence proved.

Ques. Prove that (sin α + cos α) (tan α + cot α) = sec α + cosec α. (4 marks)

Ans. We can prove the identity (sin α + cos α) (tan α + cot α) = sec α + cosec α by:

L.H.S, (sin α + cos α) (tan α + cot α)

= sin α tan α + sin α cot α + cos α tan α + cos α cot α … (by applying the distributive property)

= (sin α/cos α) + (cos α/sin α) + (sin α/cos α) + (cos α/sin α) … (with the definitions of tangent and cotangent)

= (sin² α + cos² α)/(sin α cos α) + (cos² α + sin² α)/(sin α cos α) … (using cross-multiplication)

= 2/(sin α cos α) (by using the identity sin² α + cos² α = 1)

R.H.S, sec α + cosec α

= 1/cos α + 1/sin α … (with the definitions of secant and cosecant)

Here 1/cos α = sec α and 1/ sin α = cosec α.

= (sin α + cos α)/(sin α cos α) … (using cross-multiplication)

It can be seen that the LHS is equal to 2/(sin α cos α) and the RHS is equal to (sin α + cos α)/(sin α cos α), meaning that they are equal to each other.

Hence proved.

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Multiple Angle Formulas: Sine, Cosine & Tangent

What are Multiple Angles?