MULTIPLE AND SUB-MULTIPLE ANGLES (Excercise)

\[1.\ Prove\ that\ \frac{Sin\ 3\ \theta}{Sin\ \theta}\ -\ \frac{Cos\ 3\ \theta}{Cos\ \theta}\ =\ 2\ \hspace{15cm}\]
\[\color {black}{Solution:}\ L.\ H.\ S\ =\ \frac{Sin\ 3\ \theta}{Sin\ \theta}\ -\ \frac{Cos\ 3\ \theta}{Cos\ \theta}\ \hspace{18cm}\]
\[=\ \frac{3\ Sin\ \theta\ -\ 4\ Sin^3\ \theta}{Sin\ \theta}\ -\ \frac{4\ Cos^3\ \theta\ -\ 3\ Cos\ \theta}{Cos\ \theta}\ \hspace{15cm}\]
\[=\ \frac{Sin\ \theta(3\ -\ 4\ Sin^2\ \theta)}{Sin\ \theta}\ -\ \frac{Cos\ \theta(4\ Cos^2\ \theta\ -\ 3)}{Cos\ \theta}\ \hspace{10cm}\]
\[=\ 3\ -\ 4\ Sin^2\ \theta\ -\ (4\ Cos^2\ \theta\ -\ 3)\ \hspace{10cm}\]
\[=\ 3\ -\ 4\ Sin^2\ \theta\ -\ 4\ Cos^2\ \theta\ +\ 3\ \hspace{10cm}\]
\[=\ 6\ -\ 4(Sin^2\ \theta\ +\ Cos^2\ \theta)\ \hspace{10cm}\]
\[=\ 6\ -\ 4(1)\ \hspace{10cm}\]
\[=\ 6\ -\ 4\ \hspace{10cm}\]
\[\frac{Sin\ 3\ \theta}{Sin\ \theta}\ -\ \frac{Cos\ 3\ \theta}{Cos\ \theta}\ =\ 2\ \hspace{10cm}\]