# TRIGONOMETRIC IDENTITIES (REVISION)

$1.\ If\ Tan\ A\ =\ \frac{5}{6} \ and\ Tan\ B\ =\ \frac{1}{11},\ Show\ that\ A\ +\ B\ =\ \frac{π}{4}\ \hspace{15cm}$
$\color {black}{Solution:}\ Given\ Tan\ A\ =\ \frac{5}{6} \ and\ Tan\ B\ =\ \frac{1}{11}\ \hspace{18cm}$
$W.\ K.\ T\ Tan(A + B)\ =\ \frac{Tan A\ +\ Tan B}{1\ -\ Tan A\ Tan B}\ \hspace{15cm}$
$=\ \frac{\frac{5}{6}\ +\ \frac{1}{11}}{1\ -\ \frac{5}{6}\ ×\ \frac{1}{11}}\ \hspace{10cm}$
$=\ \frac{\frac{55\ +\ 6}{66}}{1\ -\ \frac{5}{66}}\ \hspace{10cm}$
$=\ \frac{\frac{61}{66}}{\frac{66\ -\ 5}{66}}\ \hspace{10cm}$
$=\ \frac{\frac{61}{66}}{\frac{61}{66}}\ \hspace{10cm}$
$=\ 1\ \hspace{10cm}$
$Tan\ (A\ +\ B)\ =\ 1\ \hspace{10cm}$
$\implies\ (A\ +\ B)\ =\ Tan\ ^{-1}\ (1)\ \hspace{10cm}$
$\implies\ (A\ +\ B)\ =\ \frac{π}{4}\ \hspace{10cm}$
$2.\ If\ Cos\ A\ =\ \frac{1}{7} \ and\ Cos\ B\ =\ \frac{13}{14},\ Prove\ that\ A\ -\ B\ =\ \frac{π}{3}\ \hspace{15cm}$
$\color {black}{Solution:}\ Given\ Cos\ A\ =\ \frac{1}{7} \ and\ Cos\ B\ =\ \frac{13}{14}\ \hspace{18cm}$
$W.\ K.\ T\ Cos( A\ -\ B)\ =\ Cos A\ Cos B\ +\ Sin A\ Sin B\ \hspace{15cm}$
$Sin\ A\ =\ ?\ ,\ Sin\ B\ = ?\ \hspace{10cm}$
$Sin\ A\ =\ \sqrt{1\ -\ Cos^2\ A}\ \hspace{10cm}$
$=\ \sqrt{1\ -\ (\frac{1}{7})^2}\ \hspace{10cm}$
$=\ \sqrt{1\ -\ \frac{1}{49}}\ \hspace{10cm}$
$=\ \sqrt{\frac{49\ -\ 1}{49}}\ \hspace{10cm}$
$=\ \sqrt{\frac{48}{49}}\ \hspace{10cm}$
$=\ \frac{\sqrt{48}}{7}\ \hspace{10cm}$
$Sin\ A\ =\ \frac{4\ \sqrt{3}}{7}\ \hspace{10cm}$
$Sin\ B\ =\ \sqrt{1\ -\ Cos^2\ B}\ \hspace{10cm}$
$=\ \sqrt{1\ -\ (\frac{13}{14})^2}\ \hspace{10cm}$
$=\ \sqrt{1\ -\ \frac{169}{196}}\ \hspace{10cm}$
$=\ \sqrt{\frac{196\ -\ 169}{196}}\ \hspace{10cm}$
$=\ \sqrt{\frac{27}{196}}\ \hspace{10cm}$
$=\ \frac{\sqrt{27}}{14}\ \hspace{10cm}$
$Sin\ B\ =\ \frac{3\ \sqrt{3}}{14}\ \hspace{10cm}$
$Cos( A\ -\ B)\ =\ (\frac{1}{7})\ (\frac{13}{14})\ +\ (\frac{4\ \sqrt{3}}{7}\ (\frac{3\ \sqrt{3}}{14})\ \hspace{10cm}$
$=\ \frac{13}{98}\ +\ \frac{12\ (3)}{98}\ \hspace{10cm}$
$=\ \frac{13}{98}\ +\ \frac{36}{98}\ \hspace{10cm}$
$=\ \frac{13\ +\ 36}{98}\ \hspace{10cm}$
$=\ \frac{49}{98}\ \hspace{10cm}$
$Cos( A\ -\ B)\ =\ \frac{1}{2}\ \hspace{10cm}$
$(A\ -\ B)\ =\ \frac{π}{3}\ \hspace{10cm}$
$3.\ If\ Sin\ A\ =\ \frac{3}{5},\ find\ Sin\ 3\ A\ \hspace{15cm}$
$\color {black}{Solution:}\ Given\ Sin\ A\ =\ \frac{3}{5}\ \hspace{18cm}$
$W.\ K.\ T\ Sin\ 3\ A\ =\ 3\ Sin\ A\ -\ 4\ Sin^3\ A\ \hspace{10cm}$
$=\ 3( \frac{3}{5})\ -\ 4(\frac{3}{5})^3\ \hspace{10cm}$
$=\ \frac{9}{5}\ -\ \frac{108}{125}\ \hspace{10cm}$
$=\ \frac{225\ -\ 108}{125}\ \hspace{10cm}$
$=\ \frac{117}{125}\ \hspace{10cm}$
$4.\ Prove\ that\ \frac{Sin\ 3A}{Sin\ A}\ -\ \frac{Cos\ 3A}{Cos\ A}\ =\ 2\ \hspace{15cm}$
$\color {black}{Solution:}\ L.\ H.\ S\ =\ \frac{Sin\ 3A}{Sin\ A}\ -\ \frac{Cos\ 3A}{Cos\ A}\ \hspace{18cm}$
$=\ \frac{3\ Sin\ A\ -\ 4\ Sin^3\ A}{Sin\ A}\ -\ \frac{4\ Cos^3\ A\ -\ 3\ Cos\ A}{Cos\ A}\ \hspace{15cm}$
$=\ \frac{Sin\ A(3\ -\ 4\ Sin^2\ A)}{Sin\ A}\ -\ \frac{Cos\ A(4\ Cos^2\ A\ -\ 3)}{Cos\ A}\ \hspace{10cm}$
$=\ 3\ -\ 4\ Sin^2\ A\ -\ (4\ Cos^2\ A\ -\ 3)\ \hspace{10cm}$
$=\ 3\ -\ 4\ Sin^2\ A\ -\ 4\ Cos^2\ A\ +\ 3\ \hspace{10cm}$
$=\ 6\ -\ 4(Sin^2\ A\ +\ Cos^2\ A)\ \hspace{10cm}$
$=\ 6\ -\ 4(1)\ \hspace{10cm}$
$=\ 6\ -\ 4\ \hspace{10cm}$
$\frac{Sin\ 3A}{Sin\ A}\ -\ \frac{Cos\ 3A}{Cos\ A}\ =\ 2\ \hspace{10cm}$