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) $\sin 3A=3\sin A-4{\sin }^{3}A$

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

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.
$\begin{array}{rl}1\cdot \sin 3A& =\sin (2A+A)=\sin 2A\cos A+\cos 2A\sin A\\ & =2\sin A\cos A\cdot \cos A+(1-2{\sin }^{2}A)\sin A\\ & =2\sin A{\cos }^{2}A+\sin A-2{\sin }^{3}A\\ & =2\sin A(1-{\sin }^{2}A)+\sin A-2{\sin }^{3}A\\ & =2\sin A-2{\sin }^{3}A+\sin A-2{\sin }^{3}A\\ & =3\sin A-4{\sin }^{3}A\end{array}$

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

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

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

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

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

$\begin{array}{rl}=& \frac{\tan \mathrm{A}(3{\sec }^2\mathrm{A}-4{\tan }^2\mathrm{A})}{4-3{\sec }^2\mathrm{A}}\\ =& \frac{\tan \mathrm{A}(3+3{\tan }^2\mathrm{A}-4{\tan }^2\mathrm{A})}{4-3-3{\tan }^2\mathrm{A}}\\ & \tan \mathrm{A}(3-{\tan }^2\mathrm{A})3\tan \mathrm{A}-{\tan }^3\mathrm{A}\end{array}$

Recommended Video Based on Triple Angle Formula:

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 $$
[JEE MAINS 2022]

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

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

$\begin{array}{c}16\sin {20}^{\circ }\sin {40}^{\circ }\sin {80}^{\circ }\\ 16\frac{\sin {60}^{\circ }}{4}=4\left(\frac{\sqrt{3}}{2}\right)=2\sqrt{3}\end{array}$
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 $\mathrm{△}le$
$\theta$ is the smallest angle
Three angles are, $\theta ,\pi -3\theta ,2\theta$.
Given, $2b=a+c$
Use the sine rule.

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

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

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

Where $A$ is the angle

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

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

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

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