# TRIGONOMETRIC IDENTITIES (Excercise Problems)

$\LARGE{\color {purple} {PART- A}}$
$\color {purple} {1.}\ \color {red} {Find\ the\ value\ of}\ sin\ {72}^0\ cos\ {18}^0\ +\ cos\ {72}^0\ sin\ {18}^0\ \hspace{15cm}$
$\color {blue}{Solution:}\ W.\ K.\ T\ Sin ( A + B )\ =\ Sin A\ Cos B\ +\ Cos A\ Sin B\ \hspace{18cm}$
$sin\ {72}^0\ cos\ {18}^0\ +\ cos\ {72}^0\ sin\ {18}^0\ =\ Sin\ ({72}^0\ + {18}^0)\ \hspace{10cm}$
$=\ Sin \ {90}^0\ \hspace{10cm}$
$=\ 1\ \hspace{10cm}$
$\boxed {sin\ {72}^0\ cos\ {18}^0\ +\ cos\ {72}^0\ sin\ {18}^0\ =\ 1}$
$\color {purple} {2.}\ \color {red} {Find\ the\ value\ of}\ cos\ {50}^0\ cos\ {40}^0\ -\ sin\ {50}^0\ sin\ {40}^0\ \hspace{15cm}$
$\color {blue}{Solution:}\ W.\ K.\ T\ Cos A\ Cos B\ -\ Sin A Sin B\ =\ Cos( A + B )\ \hspace{18cm}$
$cos\ {50}^0\ cos\ {40}^0\ -\ sin\ {50}^0\ sin\ {40}^0\ =\ cos\ ({50}^0\ + {40}^0)\ \hspace{10cm}$
$=\ Cos\ {90}^0\ \hspace{10cm}$
$=\ 0\ \hspace{10cm}$
$\color {purple} {3.}\ \color {red} {Find\ the\ value\ of}\ \frac{Tan\ {23}^0\ +\ Tan\ {22}^0}{1\ -\ Tan\ {23}^0\ Tan\ {22}^0}\ \hspace{15cm}$
$\color {blue}{Solution:}\ W.\ K.\ T\ \frac{Tan A\ +\ Tan B}{1\ -\ Tan A\ Tan B}\ =\ Tan(A + B)\ \hspace{18cm}$
$\frac{Tan\ {23}^0\ +\ Tan\ {22}^0}{1\ -\ Tan\ {23}^0\ Tan\ {22}^0}\ =\ Tan({23}^0\ +\ {22}^0)\ =\ Tan{45}^0\ =\ 1\ \hspace{10cm}$
$\color {purple} {4.}\ \color {red} {Find\ the\ value\ of}\ \frac{Tan\ {80}^0\ -\ Tan\ {20}^0}{1\ +\ Tan\ {80}^0\ Tan\ {20}^0}\ \hspace{15cm}$
$\color {blue}{Solution:}\ W.\ K.\ T\ \frac{Tan A\ -\ Tan B}{1\ +\ Tan A\ Tan B}\ =\ Tan(A – B)\ \hspace{18cm}$
$\frac{Tan\ {80}^0\ -\ Tan\ {20}^0}{1\ +\ Tan\ {80}^0\ Tan\ {20}^0}\ =\ Tan({80}^0\ -\ {20}^0)\ =\ Tan{60}^0\ =\ \sqrt{3}\ \hspace{10cm}$
$\color {purple} {5.}\ \color {red} {Find\ the\ value\ of}\ 2\ Sin\ 15^{0}\ Cos\ 15^{0}\ \hspace{15cm}$
$\color {blue}{Solution:}\ 2\ Sin\ 15^{0}\ Cos\ 15^{0}\ = \hspace{18cm}$
$=\ Sin\ 2(15^{0})\ \hspace{15cm}$
$=\ Sin\ 30^{0}\ \hspace{15cm}$
$=\ \frac{\sqrt{1}}{2}\ \hspace{15cm}$
$\color {purple} {6.}\ \color {red} {Find\ the\ value\ of}\ 1\ -\ 2\ Sin^2\ 22\ \frac{1}{2}^{0}\ \hspace{15cm}$
$\color {blue}{Solution:}\ 1\ -\ 2\ Sin^2\ 22\ \frac{1}{2}^{0}\ =\ \hspace{18cm}$
$=\ Cos\ 2(22\ \frac{1}{2}^{0})\ \hspace{2cm}\ \because\ 1\ -\ 2\ Sin^2\ A\ =\ Cos\ 2A$
$=\ Cos\ 2(\frac{45}{2}^0)\ hspace{10cm}$
$=\ Cos\ 45^{0}\ \hspace{15cm}$
$=\ \frac{1}{\sqrt{2}}\ \hspace{10cm}$
$\color {purple} {7.}\ \color {red} {Find\ the\ value\ of}\ 3\ Sin\ 20^{0}\ -\ 4\ Sin^3\ 20^{0}\ \hspace{15cm}$
$\color {blue}{Solution:}\ 3\ Sin\ 20^{0}\ -\ 4\ Sin^3\ 20^{0}\ = \hspace{18cm}$
$=\ Sin\ 3(20^{0})\ \hspace{2cm}\ \because\ Sin\ 3A\ =\ 3\ Sin\ A\ -\ 4\ Sin^3\ A$
$=\ Sin\ 60^{0}\ \hspace{15cm}$
$=\ \frac{\sqrt{3}}{2}\ \hspace{15cm}$
$\color {purple} {8.}\ \color {red} {Find\ the\ value\ of}\ 4\ Cos^3\ 20^{0}\ -\ 3\ Cos\ 20^{0}\ \hspace{15cm}$
$\color {blue}{Solution:}\ 4\ Cos^3\ 20^{0}\ -\ 3\ Cos\ 20^{0}\ = \hspace{18cm}$
$=\ Cos\ 3(20^{0})\ \hspace{2cm}\ \because\ Cos\ 3A\ =\ 4\ Cos^3\ A\ -\ 3\ Cos\ A$
$=\ Cos\ 60^{0}\ \hspace{15cm}$
$=\ \frac{1}{2}\ \hspace{15cm}$
$\LARGE{\color {purple} {PART- B}}$
$\color {purple} {9.}\ \color {red} {Show\ that}\ \frac{Sin\ 2A}{1\ -\ Cos\ 2A}\ =\ Cot\ A\ \hspace{15cm}$
$\color {blue}{Solution:}\ LHS\ =\ \frac{Sin\ 2A}{1\ -\ Cos\ 2A}\ \hspace{18cm}$
$=\ \frac{2\ Sin\ A\ Cos\ A}{2\ Sin^2\ A}\ \hspace{2cm}\ W.\ K.\ T.\ Sin\ 2A\ =\ 2\ Sin\ A\ Cos\ A\ and\ 1\ -\ Cos\ 2A\ =\ 2\ Sin^2\ A$
$=\ \frac{Cos\ A}{Sin\ A}\ \hspace{15cm}$
$=\ Cot\ A\ =\ R.H.S\ \hspace{15cm}$
$\color {purple} {10.}\ If\ Sin\ θ\ =\ \frac{2}{3},\ \color {red} {find\ Sin\ 3θ}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Given\ Sin\ θ\ =\ \frac{2}{3}\ \hspace{18cm}$
$W.\ K.\ T\ Sin\ 3\ \theta\ =\ 3\ Sin\ \theta\ -\ 4\ Sin^3\ \theta\ \hspace{10cm}$
$=\ 3( \frac{2}{3})\ -\ 4(\frac{2}{3})^3\ \hspace{10cm}$
$=\ 2\ -\ \frac{32}{27}\ \hspace{10cm}$
$=\ \frac{54\ -\ 32}{27}\ \hspace{10cm}$
$=\ \frac{22}{27}\ \hspace{10cm}$
$\color {purple} {11.}\ If\ Sin\ θ\ =\ \frac{3}{5},\ \color {red} {find\ Sin\ 3θ}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Given\ Sin\ θ\ =\ \frac{3}{5}\ \hspace{18cm}$
$W.\ K.\ T\ Sin\ 3\ \theta\ =\ 3\ Sin\ \theta\ -\ 4\ Sin^3\ \theta\ \hspace{10cm}$
$=\ 3( \frac{3}{5})\ -\ 4(\frac{3}{5})^3\ \hspace{10cm}$
$=\ \frac{9}{5}\ -\ \frac{108}{125}\ \hspace{10cm}$
$=\ \frac{225\ -\ 108}{125}\ \hspace{10cm}$
$=\ \frac{117}{125}\ \hspace{10cm}$
$\LARGE{\color {purple} {PART- C}}$
$\color {purple} {12:}\ If\ Sin\ A\ =\ \frac{8}{17} \ and\ Sin\ B\ =\ \frac{5}{13},\ \color {red} {Show\ that\ Sin(A\ +\ B)\ =\ \frac{171}{221}}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Given\ Sin\ A\ =\ \frac{8}{17} \ and\ Sin\ B\ =\ \frac{5}{13}\ \hspace{18cm}$
$W.\ K.\ T\ Sin ( A + B )\ =\ Sin A\ Cos B\ +\ Cos A\ Sin B\ \hspace{15cm}$
$Cos\ A\ =\ ?\ ,\ Cos\ B\ = ?\ \hspace{10cm}$
$Cos\ A\ =\ \sqrt{1\ -\ Sin^2\ A}\ \hspace{10cm}$
$=\ \sqrt{1\ -\ (\frac{8}{17})^2}\ \hspace{10cm}$
$=\ \sqrt{1\ -\ \frac{64}{289}}\ \hspace{10cm}$
$=\ \sqrt{\frac{289\ -\ 64}{289}}\ \hspace{10cm}$
$=\ \sqrt{\frac{225}{289}}\ \hspace{10cm}$
$Cos\ A\ =\ \frac{15}{17}\ \hspace{10cm}$
$Cos\ B\ =\ \sqrt{1\ -\ Sin^2\ B}\ \hspace{10cm}$
$=\ \sqrt{1\ -\ (\frac{5}{13})^2}\ \hspace{10cm}$
$=\ \sqrt{1\ -\ \frac{25}{169}}\ \hspace{10cm}$
$=\ \sqrt{\frac{169\ -\ 25}{169}}\ \hspace{10cm}$
$=\ \sqrt{\frac{144}{169}}\ \hspace{10cm}$
$Cos\ B\ =\ \frac{12}{13}\ \hspace{10cm}$
$Sin ( A + B )\ =\ (\frac{8}{17})\ (\frac{12}{13})\ +\ (\frac{15}{17})\ (\frac{5}{13})\ \hspace{10cm}$
$=\ \frac{96}{221}\ +\ \frac{75}{221}\ \hspace{10cm}$
$=\ \frac{96\ +\ 75}{221}\ \hspace{10cm}$
$=\ \frac{171}{221}\ \hspace{10cm}$
$\boxed {\therefore\ Sin ( A + B )\ =\ \frac{171}{221}}\ \hspace{10cm}$