Triple Angle Identities - Practice Questions & MCQ

Triple Angle Formula
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. $\tan 3 \mathrm{~A}=\frac{3 \tan \mathrm{~A}-\tan ^3 \mathrm{~A}}{1-3 \tan ^2 \mathrm{~A}}$

Proof:
These formulas can be derived from the addition formulas and double angle formula. For example, 3 A can be written as $(2 \mathrm{~A}+\mathrm{A})$ and then apply addition formula and double angle formulas to get the results.

$
\text { 1. } \begin{aligned}
\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}
$

$
\text { 2. } \begin{aligned}
\cos 3 A & =\cos (2 A+A)=\cos 2 A \cdot A \cos A-\sin 2 A \sin A \\
& =\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.

$
\begin{aligned}
\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}
\end{aligned}
$
On dividing 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}} \\
& =\frac{\tan \mathrm{A}\left(3-\tan ^2 \mathrm{~A}\right)}{1-3 \tan ^2 \mathrm{~A}}=\frac{3 \tan \mathrm{~A}-\tan ^3 \mathrm{~A}}{1-3 \tan ^2 \mathrm{~A}}
\end{aligned}
$