# TRIGONOMETRIC IDENTITIES (Text)

$\color {royalblue} {Introduction}:\ \hspace{20cm}$
$Trigonometry\ is\ one\ of\ the\ oldest\ branches\ of\ Mathematics.\ The\ word$
$Trigonometry\ is\ derived\ from\ the\ Greek\ words\ ‘Trigonon’\ and\ ‘metron’\ means$
$measurement\ of\ angles.\ In\ olden\ days\ Trigonometry\ was\ mainly\ used\ as\ a\ tool\ for$
$studying\ astronomy.\ In\ earlier\ stages\ Trigonometry\ was\ mainly\ concerned\ with\ angles$
$of\ a\ triangle.\ But\ now\ it\ has \ its\ applications\ in\ various\ branches\ of\ science\ such\ as$
$surveying,\ engineering,\ navigations\ etc.\ For\ the\ study\ of\ higher\ mathematics,\ knowledge\ of\ Trigonometry\ is\ essential.$
$\color {royalblue} {Trigonometrical\ ratios}:\ \hspace{20cm}$
$There\ are\ six\ Trigonometrical\ ratios\ sine,\ cosine,\ tangent,\ cotangent,\ secant\ and$$cosecant\ shortly\ written\ as\ sin θ,\ cos θ,\ tan θ,\ cot θ,\ sec θ\ and\ cosec θ.$
$\color {royalblue} {Fundamental\ Trigonometrical\ identities}:\ \hspace{20cm}$
$1\ .\ sin^2θ\ +\ cos^2θ\ =\ 1\ \hspace{10cm}$
$2\ .\ 1\ +\ tan^2θ\ =\ sec^2θ\ \hspace{10cm}$
$3\ .\ 1\ +\ cot^2θ\ =\ cosec^2θ\ \hspace{10cm}$
$\color {royalblue} {Trigonometrical\ ratios\ of\ known\ angles}:\ \hspace{20cm}$
$\color {royalblue} {Signs\ of\ Trigonometrical\ ratios}:\ \hspace{20cm}$
$\color {brown} {Working\ rule\ for\ 180 ± θ\ and\ 360 ± θ}:\ \hspace{20cm}$
$\color {brown} {Working\ rule\ for\ 90 ± θ}:\ \hspace{20cm}$
$\color {purple} {Example\ 1:}\ \color {red} {Show\ that}\ tan (765^0)\ =\ 1\ \hspace{15cm}$
$\color {blue}{Solution:}\ \hspace{20cm}$
$tan(765^0)\ =\ tan(360^0\ +\ 405^0)\ \hspace{18cm}$
$=\ tan(405^0)\ \hspace{18cm}$
$=\ tan(360^0\ +\ 45^0)\ \hspace{18cm}$
$=\ tan(45^0)\ \hspace{18cm}$
$=\ 1\ \hspace{18cm}$
$\boxed{Hence,\ tan (765^0)\ =\ 1}$
$\color {royalblue} {Formulae}:\ \hspace{20cm}$
$1)\ Sin ( A + B )\ =\ Sin A\ Cos B\ +\ Cos A\ Sin B$
$2)\ Sin ( A – B )\ =\ Sin A\ Cos B\ -\ Cos A\ Sin B$
$3)\ Cos( A + B )\ =\ Cos A\ Cos B\ -\ Sin A Sin B$
$4)\ Cos( A – B )\ =\ Cos A\ Cos B\ +\ Sin A Sin B$
$5)\ Tan(A + B)\ =\ \frac{Tan A\ +\ Tan B}{1\ -\ Tan A\ Tan B}$
$6)\ Tan(A – B)\ =\ \frac{Tan A\ -\ Tan B}{1\ +\ Tan A\ Tan B}$
$\color {brown} {Note}:\ Sin\ ( – θ )\ =\ -\ sin\ θ\ \hspace{2cm}\ cos\ ( – θ )\ =\ cos\ θ$
$\color {purple} {Example\ 2:}\ \color {red} {Find\ the\ value\ of}\ sin\ {50}^0\ cos\ {40}^0\ +\ cos\ {50}^0\ sin\ {40}^0\ \hspace{15cm}$
$\color {blue}{Solution:}\ W.\ K.\ T\ Sin ( A + B )\ =\ Sin A\ Cos B\ +\ Cos A\ Sin B\ \hspace{18cm}$
$sin\ {50}^0\ cos\ {40}^0\ +\ cos\ {50}^0\ sin\ {40}^0\ =\ Sin\ ({50}^0\ + {40}^0)\ \hspace{10cm}$
$=\ Sin \ {90}^0\ \hspace{10cm}$
$=\ 1\ \hspace{10cm}$
$\boxed{\therefore\ sin\ {50}^0\ cos\ {40}^0\ +\ cos\ {50}^0\ sin\ {40}^0\ =\ 1}$
$\color {purple} {Example\ 3:}\ \color {red} {Find\ the\ value\ of}\ cos\ {40}^0\ cos\ {20}^0\ -\ sin\ {40}^0\ sin\ {20}^0\ \hspace{15cm}$
$\color {blue}{Solution:}\ W.\ K.\ T\ Cos A\ Cos B\ -\ Sin A Sin B\ =\ Cos( A + B )\ \hspace{18cm}$
$cos\ {40}^0\ cos\ {20}^0\ -\ sin\ {40}^0\ sin\ {20}^0\ =\ cos\ ({40}^0\ + {20}^0)\ \hspace{10cm}$
$=\ Cos\ {60}^0\ \hspace{10cm}$
$=\ \frac{1}{2}\ \hspace{10cm}$
$\boxed{\therefore\ cos\ {40}^0\ cos\ {20}^0\ -\ sin\ {40}^0\ sin\ {20}^0\ =\ \frac{1}{2}}$
$\color {purple} {Example\ 4:}\ \color {red} {Find\ the\ value\ of}\ \frac{Tan\ {20}^0\ +\ Tan\ {25}^0}{1\ -\ Tan\ {20}^0\ Tan\ {25}^0}\ \hspace{15cm}$
$\color {blue}{Solution:}\ W.\ K.\ T\ \frac{Tan A\ +\ Tan B}{1\ -\ Tan A\ Tan B}\ =\ Tan(A + B)\ \hspace{18cm}$
$\frac{Tan\ {20}^0\ +\ Tan\ {25}^0}{1\ -\ Tan\ {20}^0\ Tan\ {25}^0}\ =\ Tan({20}^0\ +\ {25}^0)\ =\ Tan{45}^0\ =\ 1\ \hspace{10cm}$
$\color {purple} {Example\ 5:}\ If\ Tan\ A\ =\ \frac{10}{11} \ and\ Tan\ B\ =\ \frac{1}{21},\ \color {red} {Show\ that\ A\ +\ B\ =\ {45}^0}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Given\ Tan\ A\ =\ \frac{10}{11} \ and\ Tan\ B\ =\ \frac{1}{21}\ \hspace{18cm}$
$W.\ K.\ T\ Tan(A + B)\ =\ \frac{Tan A\ +\ Tan B}{1\ -\ Tan A\ Tan B}\ \hspace{15cm}$
$=\ \frac{\frac{10}{11}\ +\ \frac{1}{21}}{1\ -\ \frac{10}{11}\ ×\ \frac{1}{21}}\ \hspace{10cm}$
$=\ \frac{\frac{210\ +\ 11}{231}}{1\ -\ \frac{10}{231}}\ \hspace{10cm}$
$=\ \frac{\frac{221}{231}}{\frac{231\ -\ 10}{231}}\ \hspace{10cm}$
$=\ \frac{\frac{221}{231}}{\frac{221}{231}}\ \hspace{10cm}$
$=\ 1\ \hspace{10cm}$
$Tan\ (A\ +\ B)\ =\ 1\ \hspace{10cm}$
$\implies\ (A\ +\ B)\ =\ Tan\ ^{-1}\ (1)\ \hspace{10cm}$
$\implies\ (A\ +\ B)\ =\ {45}^0\ \hspace{10cm}$
$\color {purple} {Example\ 6:}\ If\ Sin\ A\ =\ \frac{3}{5} \ and\ Cos\ B\ =\ \frac{12}{13},\ \color {red} {Prove\ that\ Sin(A\ +\ B)\ =\ \frac{56}{65}}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Given\ Sin\ A\ =\ \frac{3}{5} \ and\ Cos\ B\ =\ \frac{12}{13}\ \hspace{18cm}$
$W.\ K.\ T\ Sin ( A + B )\ =\ Sin A\ Cos B\ +\ Cos A\ Sin B\ \hspace{15cm}$
$Cos\ A\ =\ ?\ ,\ Sin\ B\ = ?\ \hspace{10cm}$
$Cos\ A\ =\ \sqrt{1\ -\ Sin^2\ A}\ \hspace{10cm}$
$=\ \sqrt{1\ -\ (\frac{3}{5})^2}\ \hspace{10cm}$
$=\ \sqrt{1\ -\ \frac{9}{25}}\ \hspace{10cm}$
$=\ \sqrt{\frac{25\ -\ 9}{25}}\ \hspace{10cm}$
$=\ \sqrt{\frac{16}{25}}\ \hspace{10cm}$
$Cos\ A\ =\ \frac{4}{5}\ \hspace{10cm}$
$Cos\ B\ =\ \sqrt{1\ -\ Sin^2\ B}\ \hspace{10cm}$
$=\ \sqrt{1\ -\ (\frac{12}{13})^2}\ \hspace{10cm}$
$=\ \sqrt{1\ -\ \frac{144}{169}}\ \hspace{10cm}$
$=\ \sqrt{\frac{169\ -\ 144}{169}}\ \hspace{10cm}$
$=\ \sqrt{\frac{25}{169}}\ \hspace{10cm}$
$Cos\ B\ =\ \frac{5}{13}\ \hspace{10cm}$
$Sin ( A + B )\ =\ (\frac{3}{5})\ (\frac{12}{13})\ +\ (\frac{4}{5})\ (\frac{5}{13})\ \hspace{10cm}$
$=\ \frac{36}{25}\ +\ \frac{20}{65}\ \hspace{10cm}$
$=\ \frac{36\ +\ 20}{65}\ \hspace{10cm}$
$=\ \frac{56}{65}\ \hspace{10cm}$
$\boxed {\therefore\ Sin ( A + B )\ =\ \frac{56}{65}}\ \hspace{10cm}$

$\color {purple} {Example\ 7:}\ If\ A\ and\ B\ are\ acute\ angles\ and\ Sin\ A\ =\ \frac{1}{\sqrt{10}},\ Sin\ B\ =\ \frac{1}{\sqrt{5}},\ \hspace{15cm}$$\color {red} {Prove\ that\ (A\ +\ B)\ =\ \frac{π}{4}}\ \hspace{10cm}$
$\color {blue}{Solution:}\ Given\ Sin\ A\ =\ \frac{1}{\sqrt{10}} \ and\ Sin\ B\ =\ \frac{1}{\sqrt{5}}\ \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{1}{\sqrt{10}})^2}\ \hspace{10cm}$
$=\ \sqrt{1\ -\ \frac{1}{10}}\ \hspace{10cm}$
$=\ \sqrt{\frac{10\ -\ 1}{10}}\ \hspace{10cm}$
$=\ \sqrt{\frac{9}{10}}\ \hspace{10cm}$
$Cos\ A\ =\ \frac{3}{\sqrt{10}}\ \hspace{10cm}$
$Cos\ B\ =\ \sqrt{1\ -\ Sin^2\ B }\ \hspace{10cm}$
$=\ \sqrt{1\ -\ (\frac{1}{\sqrt{5}})^2}\ \hspace{10cm}$
$=\ \sqrt{1\ -\ \frac{1}{5}}\ \hspace{10cm}$
$=\ \sqrt{\frac{5\ -\ 1}{10}}\ \hspace{10cm}$
$=\ \sqrt{\frac{4}{5}}\ \hspace{10cm}$
$Cos\ B\ =\ \frac{2}{\sqrt{5}}\ \hspace{10cm}$
$Sin ( A + B )\ =\ (\frac{1}{\sqrt{10}})\ (\frac{2}{\sqrt{5}})\ +\ (\frac{3}{\sqrt{10}})\ (\frac{1}{\sqrt{5}})\ \hspace{10cm}$
$=\ \frac{2}{\sqrt{50}}\ +\ \frac{3}{\sqrt{50}}\ \hspace{10cm}$
$=\ \frac{2\ +\ 3}{\sqrt{50}}\ \hspace{10cm}$
$=\ \frac{5}{\sqrt{50}}\ \hspace{10cm}$
$=\ \frac{5}{\sqrt{25 × 2}}\ \hspace{10cm}$
$Sin ( A + B )\ =\ \frac{1}{\sqrt{2}}\ \hspace{10cm}$
$\boxed {\therefore\ (A\ +\ B)\ =\ \frac{π}{4}}\ \hspace{10cm}$
$\color {purple} {Example\ 8:}\ If\ Sin\ A\ =\ \frac{3}{5} \ and\ Cos\ B\ =\ \frac{12}{13},\ \color {red} {find\ the\ value\ of\ Cos(A\ +\ B)}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Given\ Sin\ A\ =\ \frac{3}{5} \ and\ Cos\ B\ =\ \frac{12}{13}\ \hspace{18cm}$
$W.\ K.\ T\ Cos( A + B )\ =\ Cos A\ Cos B\ -\ Sin A\ Sin B\ \hspace{15cm}$
$Cos\ A\ =\ ?\ ,\ Sin\ B\ = ?\ \hspace{10cm}$
$Cos\ A\ =\ \sqrt{1\ -\ Sin^2\ A}\ \hspace{10cm}$
$=\ \sqrt{1\ -\ (\frac{3}{5})^2}\ \hspace{10cm}$
$=\ \sqrt{1\ -\ \frac{9}{25}}\ \hspace{10cm}$
$=\ \sqrt{\frac{25\ -\ 9}{25}}\ \hspace{10cm}$
$=\ \sqrt{\frac{16}{25}}\ \hspace{10cm}$
$Cos\ A\ =\ \frac{4}{5}\ \hspace{10cm}$
$Sin\ B\ =\ \sqrt{1\ -\ Cos^2\ B}\ \hspace{10cm}$
$=\ \sqrt{1\ -\ (\frac{12}{13})^2}\ \hspace{10cm}$
$=\ \sqrt{1\ -\ \frac{144}{169}}\ \hspace{10cm}$
$=\ \sqrt{\frac{169\ -\ 144}{169}}\ \hspace{10cm}$
$=\ \sqrt{\frac{25}{169}}\ \hspace{10cm}$
$Sin\ B\ =\ \frac{5}{13}\ \hspace{10cm}$
$Cos( A + B )\ =\ (\frac{4}{5})\ (\frac{12}{13})\ -\ (\frac{3}{5})\ (\frac{5}{13})\ \hspace{10cm}$
$=\ \frac{48}{65}\ -\ \frac{15}{65}\ \hspace{10cm}$
$=\ \frac{48\ -\ 15}{65}\ \hspace{10cm}$
$=\ \frac{33}{65}\ \hspace{10cm}$
$\boxed {Cos ( A + B )\ =\ \frac{33}{65}}\ \hspace{10cm}$
$\color {purple} {Example\ 9:}\ If\ A\ +\ B\ =\ {45}^0 \ \color {red} {Prove\ that\ (1\ +\ Tan\ A)\ (1\ +\ Tan\ B)\ =\ 2}\ \hspace{15cm}$$\color {red} {Hence\ deduce\ the\ value\ of\ Tan\ \ 22}\ {\frac{1}{2}}^0\ \hspace{13cm}$
$\color {blue}{Solution:}\ Given\ A\ +\ B\ =\ 45^{0}\ \hspace{18cm}$
$Taking\ Tan\ on\ both\ sides\ \hspace{10cm}$
$Tan\ (A\ +\ B)\ =\ Tan\ 45^{0}\ \hspace{10cm}$
$\frac{Tan A\ +\ Tan B}{1\ -\ Tan A\ Tan B}\ =\ 1\ \hspace{10cm}$
$Tan\ A\ +\ Tan\ B\ =\ 1\ -\ Tan\ A\ Tan\ B\ \hspace{10cm}$
$Tan\ A\ +\ Tan\ B\ +\ Tan\ A\ Tan\ B\ =\ 1\ ————\ (1)\ \hspace{10cm}$
$L.\ H.\ S\ =\ (1\ +\ Tan\ A)\ (1\ +\ Tan\ B)\ \hspace{10cm}$
$=\ 1\ +\ Tan\ B\ +\ Tan\ A\ +\ Tan\ A\ Tan\ B\ \hspace{10cm}$
$=\ 1\ +\ 1\ \hspace{5cm}\ using\ (1)$
$=\ 2\ =\ R.\ H.\ S\ \hspace{10cm}$
$(1\ +\ Tan\ A)\ (1\ +\ Tan\ B)\ =\ 2\ \hspace{10cm}$
$Put\ B\ =\ A\ \hspace{10cm}$
$A\ +\ A\ =\ 45^{0}\ \hspace{10cm}$
$2A\ =\ 45^{0}\ \hspace{10cm}$
$A\ =\ 22\ {\frac{1}{2}}^0\ \hspace{10cm}$
$(1\ +\ Tan\ 22\ {\frac{1}{2}}^0)\ (1\ +\ Tan\ 22\ {\frac{1}{2}}^0)\ =\ 2\ \hspace{10cm}$
$(1\ +\ Tan\ 22\ {\frac{1}{2}}^0)^2\ =\ 2\ \hspace{10cm}$
$1\ +\ Tan\ 22\ {\frac{1}{2}}^0\ =\ \sqrt{2}\ \hspace{10cm}$
$\boxed {Tan\ 22\ {\frac{1}{2}}^0\ =\ \sqrt{2}\ -\ 1}\hspace{10cm}$

## MULTIPLE AND SUB-MULTIPLE ANGLES

$\color {royalblue} {Multiple\ Angles\ of\ 2A}:\ \hspace{20cm}$
$\color {brown} {Formulae}:\ \hspace{20cm}$
$1\ (i)\ Sin\ 2A\ =\ 2\ Sin\ A\ Cos\ A\ \hspace{5cm}\ (ii)\ Sin\ 2A\ =\ \frac{2\ Tan\ A}{1\ +\ Tan^2\ A}$
$2\ (i)\ Cos\ 2A\ =\ Cos^2A\ -\ Sin^2A\ \hspace{5cm}\ (ii)\ Cos\ 2A\ =\ \frac{1\ -\ Tan^2A}{1\ +\ Tan^2\ A}$
$3.\ Tan\ 2A\ =\ \frac{2\ Tan\ A}{1\ -\ Tan^2\ A}\ \hspace{10cm}$
$4.\ Sin^2A\ =\ \frac{1\ -\ Cos\ 2A}{2}\ \hspace{5cm}\ Note:\ 1\ -\ 2\ Sin^2A\ =\ Cos\ 2A$
$5.\ Cos^2A\ =\ \frac{1\ +\ Cos\ 2A}{2}\ \hspace{10cm}$
$6.\ Tan^2A\ =\ \frac{1\ -\ Cos\ 2A}{1\ +\ Cos\ 2A}\ \hspace{10cm}$
$\color {purple} {Example\ 10:}\ \color {red} {Prove\ that}\ \frac{Sin\ 2A}{1\ +\ Cos\ 2A}\ =\ Tan\ A\ \hspace{15cm}$
$\color {blue}{Solution:}\ LHS\ =\ \frac{Sin\ 2A}{1\ +\ Cos\ 2A}\ \hspace{18cm}$
$=\ \frac{2\ Sin\ A\ Cos\ A}{2\ Cos^2\ A}\ \hspace{2cm}\ W.\ K.\ T.\ Sin\ 2A\ =\ 2\ Sin\ A\ Cos\ A\ and\ 1\ +\ Cos\ 2A\ =\ 2\ Cos^2\ A$
$=\ \frac{Sin\ A}{Cos\ A}\ \hspace{15cm}$
$=\ Tan\ A\ =\ R.H.S\ \hspace{15cm}$
$\color {purple} {Example\ 11:}\ \color {red} {Find\ the\ value\ of}\ 2\ Sin\ 30^{0}\ Cos\ 30^{0}\ \hspace{15cm}$
$\color {blue}{Solution:}\ 2\ Sin\ 30^{0}\ Cos\ 30^{0}\ = \hspace{18cm}$
$=\ Sin\ 2(30^{0})\ \hspace{15cm}$
$=\ Sin\ 60^{0}\ \hspace{15cm}$
$=\ \frac{\sqrt{3}}{2}\ \hspace{15cm}$
$\boxed {2\ Sin\ 30^{0}\ Cos\ 30^{0}\ =\ \frac{\sqrt{3}}{2}}$
$\color {purple} {Example\ 12:}\ \color {red} {Find\ the\ value\ of}\ Cos^2\ 15^{0}\ -\ Sin^2\ 15^{0}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Cos^2\ 15^{0}\ -\ Sin^2\ 15^{0}\ =\ \hspace{18cm}$
$=\ Cos\ 2(15^{0})\ \hspace{15cm}$
$=\ Cos\ 30^{0}\ \hspace{15cm}$
$=\ \frac{\sqrt{3}}{2}\ \hspace{15cm}$
$\boxed {Cos^2\ 15^{0}\ -\ Sin^2\ 15^{0}\ =\ \frac{\sqrt{3}}{2}}$
$\color {purple} {Example\ 13:}\ \color {red} {Find\ the\ value\ of}\ 1\ -\ 2\ Sin^2\ 45^{0}\ \hspace{15cm}$
$\color {blue}{Solution:}\ 1\ -\ 2\ Sin^2\ 45^{0}\ =\ \hspace{18cm}$
$=\ Cos\ 2(45^{0})\ \hspace{15cm}$
$=\ Cos\ 90^{0}\ \hspace{15cm}$
$=\ 0\ \hspace{15cm}$
$\color {purple} {Example\ 14:}\ \color {red} {Show\ that}\ \frac{1\ -\ Cos\ 2A\ +\ Sin\ 2A}{1\ +\ Cos\ 2A\ +\ Sin\ 2A}\ =\ Tan\ A\ \hspace{15cm}$
$\color {blue}{Solution:}\ LHS\ =\ \frac{1\ -\ Cos\ 2A\ +\ Sin\ 2A}{1\ +\ Cos\ 2A\ +\ Sin\ 2A}\ \hspace{18cm}$
$=\ \frac{2\ Sin^2\ A\ +\ 2\ Sin\ A\ Cos\ A}{2\ Cos^2\ A\ +\ 2\ Sin\ A\ Cos\ A}\ \hspace{2cm}\ W.\ K.\ T.\ Sin\ 2A\ =\ 2\ Sin\ A\ Cos\ A\ ,\ 1\ -\ Cos\ 2A\ =\ 2\ Sin^2\ A\ and\ 1\ +\ Cos\ 2A\ =\ 2\ Cos^2\ A$
$=\ \frac{2\ Sin\ A(Sin\ A\ +\ Cos\ A)}{2\ Cos\ A(Sin\ A\ +\ Cos\ A}\ \hspace{15cm}$
$=\ \frac{Sin\ A}{CosS\ A}\ \hspace{15cm}$
$=\ tan\ A\ =\ R.H.S\ \hspace{15cm}$
$\color {purple} {Example\ 15:}\ \color {red} {Prove\ that}\ Cos^4\ A\ -\ Sin^4\ A\ =\ Cos\ 2A\ \hspace{15cm}$
$\color {blue}{Solution:}\ L.\ H.\ S\ =\ Cos^4\ A\ -\ Sin^4\ A\ \hspace{18cm}$
$=\ (Cos^2\ A)^2\ -\ (Sin^2\ A)^2\hspace{15cm}$
$=\ (Cos^2\ A\ +\ Sin^2\ A)\ (Cos^2\ A\ +\ Sin^2\ A)\hspace{10cm}$
$=\ (1)\ (Cos^2\ A\ -\ Sin^2\ A)\hspace{10cm}$
$=\ Cos\ 2A\ =\ R.H.S\ \hspace{15cm}$
$\color {purple} {Example\ 15:}\ \color {red} {Prove\ that}\ Sin^2\ A\ +\ Sin^2\ (60^0\ +\ A)\ +\ Sin^2\ (60^0\ -\ A)\ =\ \frac{3}{2}\ \hspace{15cm}$
$\color {blue}{Solution:}\ L.\ H.\ S\ =\ Sin^2\ A\ +\ Sin^2\ (60^0\ +\ A)\ +\ Sin^2\ (60^0\ -\ A)\ \hspace{16cm}$
$=\ \frac{1\ -\ Cos\ 2A}{2}\ +\ \frac{1\ -\ Cos\ 2(60^0\ +\ A)}{2}\ +\ \frac{1\ -\ Cos\ 2(60^0\ -\ A)}{2}\ \hspace{10cm}$
$=\ \frac{1}{2}\ +\ \frac{1}{2}\ +\ \frac{1}{2}\ -\ \frac{1}{2}[Cos\ 2\ A\ +\ Cos\ (120^0\ +\ 2\ A)\ +\ Cos\ (120^0\ -\ 2\ A)]\ \hspace{10cm}$
$=\ \frac{3}{2}\ -\ \frac{1}{2}[Cos\ 2\ A\ +\ 2\ Cos\ 120^0\ Cos\ 2\ A]\ \hspace{2cm}\ \because\ Cos(A\ +\ B)\ +\ Cos(A\ -\ B)\ =\ 2\ Cos\ A\ Cos\ B$
$=\ \frac{3}{2}\ -\ \frac{1}{2}[Cos\ 2\ A\ +\ 2\ (\frac{-1}{2})\ Cos\ 2\ A]\ \hspace{10cm}$
$=\ \frac{3}{2}\ -\ \frac{1}{2}[Cos\ 2\ A\ -\ Cos\ 2\ A]\ \hspace{10cm}$
$=\ \frac{3}{2}\ -\ \frac{1}{2}[0]\ \hspace{10cm}$
$=\ \frac{3}{2}\ =\ R.H.S\ \hspace{10cm}$
$\color {purple} {Example\ 16:}\ If\ Tan\ A\ =\ \frac{1}{3} \ and\ Tan\ B\ =\ \frac{1}{7},\ \color {red} {Show\ that\ 2A\ +\ B\ =\ \frac{π}{4}}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Given\ Tan\ A\ =\ \frac{1}{3} \ and\ Tan\ B\ =\ \frac{1}{7}\ \hspace{18cm}$
$Tan(A + B)\ =\ \frac{Tan A\ +\ Tan B}{1\ -\ Tan A\ Tan B}\ \hspace{15cm}$
$Tan(2A + B)\ =\ \frac{Tan 2A\ +\ Tan B}{1\ -\ Tan 2A\ Tan B}\ —————- (1)\ \hspace{15cm}$
$Tan\ 2A\ =\ \frac{2\ Tan\ A}{1\ -\ Tan^2\ A}\ \hspace{10cm}$
$=\ \frac{2\ (\frac{1}{3})}{1\ -\ (\frac{1}{3})^2}\ \hspace{10cm}$
$=\ \frac{\frac{2}{3}}{1\ -\ \frac{1}{9}}\ \hspace{10cm}$
$=\ \frac{\frac{2}{3}}{\frac{8}{9}}\ \hspace{10cm}$
$=\ \frac{2}{3}\ ×\ \frac{9}{8}\ \hspace{10cm}$
$=\ \frac{3}{4}\ \hspace{10cm}$
$Equation\ (1)\ becomes$
$Tan(2A + B)\ =\ \frac{\frac{3}{4}\ +\ \frac{1}{7}}{1\ -\ \frac{3}{4}\ ×\ \frac{1}{7}}\ \hspace{15cm}$
$=\ \frac{\frac{21\ +\ 4}{28}}{1\ -\ \frac{3}{28}}\ \hspace{10cm}$
$=\ \frac{\frac{25}{28}}{\frac{28\ -\ 3}{28}}\ \hspace{10cm}$
$=\ \frac{\frac{25}{28}}{\frac{25}{28}}\ \hspace{10cm}$
$=\ 1\ \hspace{10cm}$
$Tan\ (2A\ +\ B)\ =\ 1\ \hspace{10cm}$
$\implies\ (2A\ +\ B)\ =\ Tan\ ^{-1}\ (1)\ \hspace{10cm}$
$\boxed{\therefore\ (2A\ +\ B)\ =\ \frac{π}{4}}\ \hspace{10cm}$
$\color {royalblue} {Multiple\ Angles\ of\ 3A}:\ \hspace{20cm}$
$\color {brown} {Formulae}:\ \hspace{20cm}$
$1.\ Sin\ 3\ \theta\ =\ 3\ Sin\ \theta\ -\ 4\ Sin^3\ \theta\ \hspace{10cm}$
$2.\ Cos\ 3\ \theta\ =\ 4\ Cos^3\ \theta\ -\ 3\ Cos\ \theta\ \hspace{10cm}$
$3.\ tan\ 3\ \theta\ =\ \frac{3\ tan\ \theta\ -\ tan^3\ \theta}{1\ -\ 3\ tan^2\ \theta}\ \hspace{10cm}$
$\color {purple} {Example\ 17:}\ If\ Sin\ θ\ =\ \frac{1}{3},\ \color {red} {find\ Sin\ 3θ}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Given\ Sin\ θ\ =\ \frac{1}{3}\ \hspace{18cm}$
$W.\ K.\ T\ Sin\ 3\ \theta\ =\ 3\ Sin\ \theta\ -\ 4\ Sin^3\ \theta\ \hspace{10cm}$
$=\ 3( \frac{1}{3})\ -\ 4(\frac{1}{3})^3\ \hspace{10cm}$
$=\ 1\ -\ \frac{4}{27}\ \hspace{10cm}$
$=\ \frac{27\ -\ 4}{27}\ \hspace{10cm}$
$=\ \frac{23}{27}\ \hspace{10cm}$
$\boxed {Sin\ 3θ\ =\ \frac{23}{27}}$
$\color {purple} {Example\ 18:}\ \color {red} {Find\ the\ value\ of}\ 3\ Sin\ 10^{0}\ -\ 4\ Sin^3\ 10^{0}\ \hspace{15cm}$
$\color {blue}{Solution:}\ 3\ Sin\ 10^{0}\ -\ 4\ Sin^3\ 10^{0}\ = \hspace{18cm}$
$=\ Sin\ 3(10^{0})\ \hspace{2cm}\ \because\ Sin\ 3A\ =\ 3\ Sin\ A\ -\ 4\ Sin^3\ A$
$=\ Sin\ 30^{0}\ \hspace{15cm}$
$=\ \frac{1}{2}\ \hspace{15cm}$
$\boxed {3\ Sin\ 10^{0}\ -\ 4\ Sin^3\ 10^{0}\ =\ \frac{1}{2}}$
$\color {purple} {Example\ 19:}\ \color {red} {Find\ the\ value\ of}\ 4\ Cos^3\ 15^{0}\ -\ 3\ Cos\ 15^{0}\ \hspace{15cm}$
$\color {blue}{Solution:}\ 4\ Cos^3\ 15^{0}\ -\ 3\ Cos\ 15^{0}\ = \hspace{18cm}$
$=\ Cos\ 3(15^{0})\ \hspace{2cm}\ \because\ Cos\ 3A\ =\ 4\ Cos^3\ A\ -\ 3\ Cos\ A$
$=\ Cos\ 45^{0}\ \hspace{15cm}$
$=\ \frac{1}{\sqrt{2}}\ \hspace{15cm}$
$\boxed{4\ Cos^3\ 15^{0}\ -\ 3\ Cos\ 15^{0}\ =\ \frac{1}{\sqrt{2}}}$
$\color {purple} {Example\ 20:}\ If\ tan\ θ\ =\ 3,\ \color {red} {find\ tan\ 3θ}\ \hspace{20cm}$
$\color {blue}{Solution:}\ Given\ tan\ θ\ =\ 3\ \hspace{18cm}$
$W.\ K.\ T\ tan\ 3\ \theta\ =\ \frac{3\ Tan\ \theta\ -\ Tan^3\ \theta}{1\ -\ 3\ Tan^2\ \theta}\ \hspace{10cm}$
$=\ \frac{3(3)\ -\ (3)^3}{1\ -\ 3(3)^2}\ \hspace{10cm}$
$=\ \frac{9\ -\ 27}{1\ -\ 27}\ \hspace{10cm}$
$=\ \frac{-\ 18}{-\ 26}\ \hspace{10cm}$
$=\ \frac{9}{13}\ \hspace{10cm}$
$\boxed{tan\ 3θ\ =\ \frac{9}{13}}$
$\color {purple} {Example\ 21:}\ \color {red} {Find\ the\ value\ of}\ \frac{3\ Tan\ {20}^0\ +\ Tan^3\ {20}^0}{1\ -\ 3\ Tan^2\ {20}^0}\ \hspace{15cm}$
$\color {blue}{Solution:}\ \frac{3\ Tan\ {20}^0\ +\ Tan^3\ {20}^0}{1\ -\ 3\ Tan^2\ {20}^0}\ =\ \hspace{18cm}$
$=\ Tan\ 3(20^{0})\ \hspace{15cm}$
$=\ Tan\ 60^{0}\ \hspace{15cm}$
$=\ \sqrt{3}\ \hspace{15cm}$
$\boxed{\frac{3\ Tan\ {20}^0\ +\ Tan^3\ {20}^0}{1\ -\ 3\ Tan^2\ {20}^0}\ =\ \sqrt{3}}$
$\color {purple} {Example\ 22:}\ \color {red} {Prove\ that}\ \frac{Cos\ 3θ}{Cos\ θ}\ +\ \frac{Sin\ 3θ}{Sin\ θ}\ =\ 4\ Cos\ 2θ\ \hspace{15cm}$
$\color {blue}{Solution:}\ L.\ H.\ S\ =\ \frac{Cos\ 3θ}{Cos\ θ}\ +\ \frac{Sin\ 3θ}{Sin\ θ}\ \hspace{18cm}$
$=\ \frac{4\ Cos^3\ θ\ -\ 3\ Cos\ θ}{Cos\ θ}\ +\ \frac{3\ Sin\ θ\ -\ 4\ Sin^3\ θ}{Sin\ θ}\ \hspace{15cm}$
$=\ \frac{Cos\ θ(4\ Cos^2\ θ\ -\ 3)}{Cos\ θ}\ +\ \frac{Sin\ θ(3\ -\ 4\ Sin^2\ θ)}{Sin\ θ}\ \hspace{10cm}$
$=\ 4\ Cos^2\ θ\ -\ 3\ +\ 3\ -\ 4\ Sin^2\ θ\ \hspace{10cm}$
$=\ 4\ (Cos^2\ θ\ -\ Sin^2\ θ)\ \hspace{10cm}$
$=\ 4\ Cos\ 2θ\ =\ R.H.S\ \hspace{10cm}$

$\color {purple} {Example\ 23:}\ \color {red} {Show\ that}\ \frac{Sin\ 3A}{1\ +\ 2\ Cos\ 2A}\ =\ Sin\ A\ \hspace{15cm}$
$\color {blue}{Solution:}\ W.\ K.\ T\ Sin\ 3A\ =\ 3\ Sin\ A\ -\ 4\ Sin^3\ A\ and\ 1\ -\ 2\ Sin^2A\ =\ Cos\ 2A\ \hspace{15cm}$
$L.\ H.\ S\ =\ \frac{Sin\ 3A}{1\ +\ 2\ Cos\ 2A}\ \hspace{15cm}$
$=\frac{3\ Sin\ A\ -\ 4\ Sin^3\ A}{1\ +\ 2\ (1\ -\ 2\ Sin^2A)}\ \hspace{10cm}$
$=\frac{Sin\ A(3\ -\ 4\ Sin^2\ A)}{1\ +\ 2\ -\ 4\ Sin^2A}\ \hspace{10cm}$
$=\frac{Sin\ A(3\ -\ 4\ Sin^2\ A)}{(3\ -\ 4\ Sin^2A)}\ \hspace{10cm}$
$=\ Sin\ A\ =\ R.H.S\ \hspace{10cm}$
$Ex\ 24:\ Prove\ that\ \frac{1\ -\ Cos\ 3A}{1\ -\ Cos\ A}\ =\ (1\ +\ 2\ Cos\ A)^2\ \hspace{15cm}$
$\color {black}{Solution:}\ L.\ H.\ S\ =\ \frac{1\ -\ Cos\ 3A}{1\ -\ Cos\ A}\ \hspace{18cm}$
$=\ \frac{1\ -\ (4\ Cos^3\ A\ -\ 3\ Cos\ A)}{1\ -\ Cos\ A}\ \hspace{10cm}$
$=\ \frac{1\ -\ 4\ Cos^3\ A\ +\ 3\ Cos\ A}{1\ -\ Cos\ A}\ \hspace{10cm}$
$=\ \frac{1\ -\ 4\ Cos^3\ A\ +\ 4\ Cos\ A\ -\ Cos\ A}{1\ -\ Cos\ A}\ \hspace{10cm}$
$=\ \frac{1\ -\ Cos\ A\ -\ 4\ Cos\ A(1\ -\ Cos^2\ A)}{1\ -\ Cos\ A}\ \hspace{10cm}$
$=\ \frac{1\ -\ Cos\ A\ -\ 4\ Cos\ A(1\ -\ Cos\ A)(1\ +\ Cos\ A)}{1\ -\ Cos\ A}\ \hspace{10cm}$
$=\ 1\ +\ 4\ Cos\ A(1\ +\ Cos\ A)\ \hspace{10cm}$
$=\ 1\ +\ 4\ Cos\ A\ +\ 4\ Cos^2\ A\ \hspace{10cm}$
$=\ (1\ +\ 2\ Cos\ A)^2\ \hspace{10cm}$

### Exercise Problems

$\LARGE{\color {purple} {PART- A}}$
$\color {purple} {1.}\ \color {red} {Find\ the\ value\ of}\ sin\ {72}^0\ cos\ {18}^0\ +\ cos\ {72}^0\ sin\ {18}^0\ \hspace{15cm}$
$\color {purple} {2.}\ \color {red} {Find\ the\ value\ of}\ cos\ {50}^0\ cos\ {40}^0\ -\ sin\ {50}^0\ sin\ {40}^0\ \hspace{15cm}$
$\color {purple} {3.}\ \color {red} {Find\ the\ value\ of}\ \frac{Tan\ {23}^0\ +\ Tan\ {22}^0}{1\ -\ Tan\ {23}^0\ Tan\ {22}^0}\ \hspace{15cm}$
$\color {purple} {4.}\ \color {red} {Find\ the\ value\ of}\ \frac{Tan\ {80}^0\ -\ Tan\ {20}^0}{1\ +\ Tan\ {80}^0\ Tan\ {20}^0}\ \hspace{15cm}$
$\color {purple} {5.}\ \color {red} {Find\ the\ value\ of}\ 2\ Sin\ 15^{0}\ Cos\ 15^{0}\ \hspace{15cm}$
$\color {purple} {6.}\ \color {red} {Find\ the\ value\ of}\ 1\ -\ 2\ Sin^2\ 22\ \frac{1}{2}^{0}\ \hspace{15cm}$
$\color {purple} {7.}\ \color {red} {Find\ the\ value\ of}\ 3\ Sin\ 20^{0}\ -\ 4\ Sin^3\ 20^{0}\ \hspace{15cm}$
$\color {purple} {8.}\ \color {red} {Find\ the\ value\ of}\ 4\ Cos^3\ 20^{0}\ -\ 3\ Cos\ 20^{0}\ \hspace{15cm}$
$\LARGE{\color {purple} {PART- B}}$
$\color {purple} {9.}\ \color {red} {Show\ that}\ \frac{Sin\ 2A}{1\ -\ Cos\ 2A}\ =\ Cot\ A\ \hspace{15cm}$
$\color {purple} {10.}\ If\ Sin\ θ\ =\ \frac{2}{3},\ \color {red} {find\ Sin\ 3θ}\ \hspace{15cm}$
$\color {purple} {11.}\ If\ Sin\ θ\ =\ \frac{3}{5},\ \color {red} {find\ Sin\ 3θ}\ \hspace{15cm}$
$\LARGE{\color {purple} {PART- C}}$
$\color {purple} {12:}\ If\ Sin\ A\ =\ \frac{8}{17} \ and\ Sin\ B\ =\ \frac{5}{13},\ \color {red} {Show\ that\ Sin(A\ +\ B)\ =\ \frac{171}{221}}\ \hspace{15cm}$