MULTIPLE AND SUB-MULTIPLE ANGLES (Revision)

\[1.\ If\ Sin\ A\ =\ \frac{3}{5},\ find\ Sin\ 3\ A\ \hspace{15cm}\]
\[\color {black}{Solution:}\ Given\ Sin\ A\ =\ \frac{3}{5}\ \hspace{18cm}\]
\[Sin\ 3A\ =\ 3\ Sin\ A\ -\ 4\ Sin^3\ A\ \hspace{10cm}\]
\[=\ 3( \frac{3}{5})\ -\ 4(\frac{3}{5})^3\ \hspace{10cm}\]
\[=\ \frac{9}{5}\ -\ 4\ (\frac{27}{125})\ \hspace{10cm}\]
\[=\ \frac{9}{5}\ -\ \frac{108}{125}\ \hspace{10cm}\]
\[=\ \frac{225\ -\ 108}{125}\ \hspace{10cm}\]
\[Sin\ 3A\ =\ \frac{117}{125}\ \hspace{10cm}\]
\[2.\ Prove\ that\ \frac{Sin\ 3A}{Sin\ A}\ -\ \frac{Cos\ 3A}{Cos\ A}\ =\ 2\ \hspace{15cm}\]
\[\color {black}{Solution:}\ L.\ H.\ S\ =\ \frac{Sin\ 3A}{Sin\ A}\ -\ \frac{Cos\ 3A}{Cos\ A}\ \hspace{18cm}\]
\[=\ \frac{3\ Sin\ A\ -\ 4\ Sin^3\ A}{Sin\ A}\ -\ \frac{4\ Cos^3\ A\ -\ 3\ Cos\ A}{Cos\ A}\ \hspace{15cm}\]
\[=\ \frac{Sin\ A(3\ -\ 4\ Sin^2\ A)}{Sin\ A}\ -\ \frac{Cos\ A(4\ Cos^2\ A\ -\ 3)}{Cos\ A}\ \hspace{10cm}\]
\[=\ 3\ -\ 4\ Sin^2\ A\ -\ (4\ Cos^2\ A\ -\ 3)\ \hspace{10cm}\]
\[=\ 3\ -\ 4\ Sin^2\ A\ -\ 4\ Cos^2\ A\ +\ 3\ \hspace{10cm}\]
\[=\ 6\ -\ 4(Sin^2\ A\ +\ Cos^2\ A)\ \hspace{10cm}\]
\[=\ 6\ -\ 4(1)\ \hspace{10cm}\]
\[=\ 6\ -\ 4\ \hspace{10cm}\]
\[\frac{Sin\ 3A}{Sin\ A}\ -\ \frac{Cos\ 3A}{Cos\ A}\ =\ 2\ \hspace{10cm}\]