Triple Angle Formulas

The Triple angle formula is used to convert the trigonometric ratios of triple angles into the trigonometric ratios of single angles. The Triple angle formula can be derived using the Trigonometric ratios formula of compound angles( Putting A=B). The triple angle formula is used to solve complex trigonometric identities and convert them to single angles.

In this article, we will cover the concept of the Triple Angle Formula. This category falls under the broader category of Trigonometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more.

Triple Angle

The Triple angle formula is used to transform the trigonometric ratios of triple angles into the trigonometric ratios of single angles.

Triple Angle Formulas

The Triple angle formula is used to transform the trigonometric ratios of triple angles into the trigonometric ratios of single angles. The Triple angle formulas can be derived from the sum formulas and double angle formulas. We have triple-angle formulas of sin, cos, and tan functions. The triple angle formula is:

1. $\sin 3 \mathrm{~A}=3 \sin \mathrm{A}-4 \sin ^3 \mathrm{~A}$
2. $\cos 3 \mathrm{~A}=4 \cos ^3 \mathrm{~A}-3 \cos A$
3. $\tan 3 \mathrm{~A}=\frac{3 \tan \mathrm{A}-\tan ^3 \mathrm{~A}}{1-3 \tan ^2 \mathrm{~A}}$

What are triple angle formulas?

The Triple angle formula is used to transform the trigonometric ratios of triple angles into the trigonometric ratios of single angles.
1) $\sin 3 A=3 \sin A-4 \sin ^3 A$

This formula is used to convert the triple angle of sine into the expression of a single angle of sine functions.
2) $\cos 3 A=4 \cos ^3 A-3 \cos A$

This formula is used to convert the triple angle of cosine into the expression of a single angle of cosine functions.
3. $\tan 3 \mathrm{~A}=\frac{3 \tan \mathrm{A}-\tan ^3 \mathrm{~A}}{1-3 \tan ^2 \mathrm{~A}}$

This formula is used to convert the triple angle of tangent into the expression of a single angle of tangent functions.

Proof of Triple-angle formulas

These formulas can be derived from the addition formulas and double angle formulas. For example, 3A can be written as (2A + A), and then apply addition formula and double angle formulas to get the results.
$
\begin{aligned}
1 \cdot \sin 3 A & =\sin (2 A+A)=\sin 2 A \cos A+\cos 2 A \sin A \\
& =2 \sin A \cos A \cdot \cos A+\left(1-2 \sin ^2 A\right) \sin A \\
& =2 \sin A \cos ^2 A+\sin A-2 \sin ^3 A \\
& =2 \sin A\left(1-\sin ^2 A\right)+\sin A-2 \sin ^3 A \\
& =2 \sin A-2 \sin ^3 A+\sin A-2 \sin ^3 A \\
& =3 \sin A-4 \sin ^3 A
\end{aligned}
$

2. $\cos 3 A=\cos (2 A+A)=\cos 2 A \cdot A \cos A-\sin 2 A \sin A$

$
\begin{aligned}
& =\left(2 \cos ^2 A-1\right) \cos A-2 \sin A \cos A \cdot \sin A \\
& =2 \cos ^3 A-\cos A-2 \cos A\left(1-\cos ^2 A\right) \\
& =2 \cos ^3 A-\cos A-2 \cos A+2 \cos ^3 A \\
& =4 \cos ^3 A-3 \cos A
\end{aligned}
$

3. $\tan 3 A=\frac{\sin 3 A}{\cos 3 A}=\frac{3 \sin A-4 \sin ^3 A}{4 \cos ^3 A-3 \cos A}$

$
=\frac{\sin \mathrm{A}\left(3-4 \sin ^2 \mathrm{~A}\right)}{\cos \mathrm{A}\left(4 \cos ^2 \mathrm{~A}-3\right)}=\frac{\tan \mathrm{A}\left(3-4 \sin ^2 \mathrm{~A}\right)}{4 \cos ^2 \mathrm{~A}-3}
$

On dividing the numerator and denominator by $\cos ^2 A$,

$
\begin{aligned}
= & \frac{\tan \mathrm{A}\left(3 \sec ^2 \mathrm{~A}-4 \tan ^2 \mathrm{~A}\right)}{4-3 \sec ^2 \mathrm{~A}} \\
= & \frac{\tan \mathrm{A}\left(3+3 \tan ^2 \mathrm{~A}-4 \tan ^2 \mathrm{~A}\right)}{4-3-3 \tan ^2 \mathrm{~A}} \\
& \tan \mathrm{A}\left(3-\tan ^2 \mathrm{~A}\right) 3 \tan \mathrm{A}-\tan ^3 \mathrm{~A}
\end{aligned}
$

Summary

The Triple angle formula in trigonometry is a relationship between the trigonometric functions of an angle and those of thrice that angle. These identities simplify complex expressions and make them simpler. Their widespread application in solving equations, and deriving identities increases their importance in both theoretical and practical concepts.

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Solved Example Based on Triple Angle Formula

Example 1: If $\sin ^2\left(10^{\circ}\right) \sin \left(20^{\circ}\right) \sin \left(40^{\circ}\right) \sin \left(50^{\circ}\right) \sin \left(70^{\circ}\right)=\alpha-\frac{1}{16} \sin \left(10^{\circ}\right)$, then $16+\alpha^{-1}$ is equal to $\qquad$
[JEE MAINS 2022]

$
\begin{aligned}
& \text { Solution } \\
& \begin{array}{l}
\sin 10^{\circ}\left(\frac{1}{2} \cdot 2 \sin 20^{\circ} \cdot \sin 40^{\circ}\right) \cdot \sin 10^{\circ} \cdot \sin \left(60^{\circ}-10^{\circ}\right) \cdot \sin \left(60^{\circ}+10^{\circ}\right) \\
\sin 10^{\circ} \cdot \frac{1}{2} \cdot\left(\cos 20^{\circ}-\cos 60^{\circ}\right) \cdot \frac{1}{4} \sin 30^{\circ} \\
\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{2} \sin 10^{\circ}\left(\cos 20^{\circ}-\frac{1}{2}\right) \\
=\frac{1}{32}\left(2 \sin 10^{\circ} \cdot \cos 20^{\circ}-\sin 10^{\circ}\right) \\
=\frac{1}{32}\left(\sin 30^{\circ}-\sin 10^{\circ}-\sin 10^{\circ}\right) \\
=\frac{1}{32}\left(\frac{1}{2}-2 \sin 10^{\circ}\right) \\
=\frac{1}{64}\left(1-4 \sin 10^{\circ}\right) \\
=\frac{1}{64}-\frac{1}{16} \sin 10^{\circ}
\end{array}
\end{aligned}
$

Hence $\alpha=\frac{1}{64}$

$
16+\alpha^{-1}=80
$
Hence, the answer is 80 .
Example 2: $16 \sin \left(20^{\circ}\right) \sin \left(40^{\circ}\right) \sin \left(80^{\circ}\right)$ is equal to.
[JEE MAINS 2022]
Solution

$
\begin{gathered}
16 \sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ} \\
16 \frac{\sin 60^{\circ}}{4}=4\left(\frac{\sqrt{3}}{2}\right)=2 \sqrt{3}
\end{gathered}
$
Hence, the answer is $2 \sqrt{3}$
Example 3: If the lengths of the sides of a triangle are in A.P. and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle is :
Solution: Let $a<b<c$ be sides of $\triangle l e$
$\theta$ is the smallest angle
Three angles are, $\theta, \pi-3 \theta, 2 \theta$.
Given, $2 b=a+c$
Use the sine rule.

Hence, the answer is 4:5:6
Example 4: If $\sin A=\frac{\sqrt{3}}{2}$ and $\sin \left(60^{\circ}+A\right)=\frac{\sqrt{3}}{2}$, then find $\sin 3 A$
Solution: Results of Triple Angle Formula-

$
\sin 3 A=4 \sin \left(60^{\circ}-A\right) \cdot \sin A \cdot \sin \left(60^{\circ}+A\right)
$

$
\begin{aligned}
& 2 \sin (B)=\sin (A)+\sin (C) \\
& 2 \sin (3 \theta)=\sin (\theta)+\sin (2 \theta) \\
& 2\left(3 \sin \theta-4 \sin ^3 \theta\right)=\sin \theta(1+2 \cos \theta) \\
& 6-8\left(1-\cos ^2 \theta\right)=1+2 \cos \theta \\
& \cos \theta=\frac{3}{4},-\frac{1}{2}\left(-\frac{1}{2} \text { is rejected }\right) \\
& : a: b: c=\sin A: \sin B: \sin C \\
& =\sin \theta: \sin 3 \theta: \sin 2 \theta \\
& =1: 3-4 \sin ^2 \theta: 2 \cos \theta=1: 4 \cos ^2 \theta-1: 2 \cos \theta \\
& =4: 5: 6
\end{aligned}
$

Where $A$ is the angle

$
\begin{aligned}
& \sin 3 A=4 \sin \left(60^{\circ}-A\right) \cdot \sin A \cdot \sin \left(60^{\circ}+A\right) \\
& \text { If } \sin A=\sin \left(60^{\circ}+A\right)=\frac{\sqrt{3}}{2} \Rightarrow A=60^{\circ}
\end{aligned}
$
Thus $\sin 3 A=0$
Hence, the answer is 0

Example 5: If $\tan \left(60^{\circ}-A\right)=a ; \tan \left(60^{\circ}+A\right)=b$; and $\tan 3 A=c$; then $\tan A=$
Solution: Results of Triple Angle Formula-

$
\tan 3 A=\tan \left(60^{\circ}-A\right) \tan A \tan \left(60^{\circ}+A\right)
$
Where A is the angle.

$
c=a b \tan A \Rightarrow \tan A=\frac{c}{a b}
$
Hence, the answer is $\mathrm{c} / \mathrm{ab}$