Multiple Angle Formulas

The multiple angles generally appear in trigonometric functions. The values of multiple angles are not possible to find directly but their values can be calculated by expressing each trigonometric function in its expanded form. The trigonometric function of multiple angles is also known as the multiple angle formula. The double and triple angles formula are used under the multiple angle formulas. Sine, tangent, and cosine are the common functions that are used for the multiple angle formula.  Let us learn more about the multiple angle formulas in the upcoming sections. 

What Are the Multiple Angle Formulas?

The multiple angles generally appear in trigonometric functions. The values of multiple angles are not possible to find directly but their values can be calculated by expressing each trigonometric function in its expanded form. The following multiple angle formula identities are used in mathematics.

Formula 1: The sin formula for multiple angle is:

\(\large sin n\theta = \sum_{k=0}^{n}\;cos^{k}\theta \; sin^{n-k}\theta\; Sin\left [\frac{1}{2}\left(n-k\right)\right]\pi\)

where n=1,2,3,……

General formulas are, 

Sin2θ =2 × Cosθ.Sinθ

Sin3θ =3Sinθ - 4Sin³θ

Formula 2: The multiple angle’s Cosine formula is given below:

\(\large Cos\;n\, \theta =\sum_{k=0}^{n}cos^{k}\theta \,sin^{n-k}\theta \;cos\left [\frac{1}{2}\left(n-k\right)\pi\right]\)

where n = 1,2,3

The general formula goes as:

Cos2θ = Cos²θ – Sin²θ

Cos3θ = 4Cos³θ – 3Cosθ

Formula 3: Tangent Multiple Angles formula

\(Tan\;n\theta = \frac{sin\;n\theta}{cos\;n\theta}\)

Where n = 1,2,3....

Let us see how to use the multiple angle formulas in the following solved examples section.

Solved Examples Using Multiple Angle Formulas

Example 1: 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.

Solution:

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

Answer: Henced proved \(\frac{3 \sin\theta -4 sin^{3}\theta}{4 cos^{3}\, \theta – 3 cos\, \theta}=tan\;3\theta\).

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

Solution:

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 

Answer: Henced proved \(\frac{sin\,x+sin\,2x}{1+cos\,x+cos\,2x}=tan\,x\).