3.2 MULTIPLE AND SUB-MULTIPLE ANGLES

$\color {royalblue} {Multiple\ Angles\ of\ 2A}:\ \hspace{20cm}$
$\color {brown} {Formulae}:\ \hspace{20cm}$
$1\ (i)\ Sin\ 2A\ =\ 2\ Sin\ A\ Cos\ A\ \hspace{5cm}\ (ii)\ Sin\ 2A\ =\ \frac{2\ Tan\ A}{1\ +\ Tan^2\ A}$
$2\ (i)\ Cos\ 2A\ =\ Cos^2A\ -\ Sin^2A\ \hspace{5cm}\ (ii)\ Cos\ 2A\ =\ \frac{1\ -\ Tan^2A}{1\ +\ Tan^2\ A}$
$3.\ Tan\ 2A\ =\ \frac{2\ Tan\ A}{1\ -\ Tan^2\ A}\ \hspace{10cm}$
$4.\ Sin^2A\ =\ \frac{1\ -\ Cos\ 2A}{2}\ \hspace{5cm}\ Note:\ 1\ -\ 2\ Sin^2A\ =\ Cos\ 2A$
$5.\ Sin^2A\ =\ \frac{1\ +\ Cos\ 2A}{2}\ \hspace{10cm}$
$6.\ Tan^2A\ =\ \frac{1\ -\ Cos\ 2A}{1\ +\ Cos\ 2A}\ \hspace{10cm}$
$\color {royalblue} {Examples}:\ \hspace{20cm}$
$Ex\ 1:\ Prove\ that\ \frac{Sin\ 2A}{1\ +\ Cos\ 2A}\ =\ Tan\ 2A\ \hspace{15cm}$
$\color {black}{Solution:}\ LHS\ =\ \frac{Sin\ 2A}{1\ +\ Cos\ 2A}\ \hspace{18cm}$
$=\ \frac{2\ Sin\ A\ Cos\ A}{2\ Cos^2\ A}\ \hspace{2cm}\ W.\ K.\ T.\ Sin\ 2A\ =\ 2\ Sin\ A\ Cos\ A\ and\ 1\ +\ Cos\ 2A\ =\ 2\ Cos^2\ A$
$=\ \frac{Sin\ A}{Cos\ A}\ \hspace{15cm}$
$=\ Tan\ A\ =\ R.H.S\ \hspace{15cm}$
$Ex\ 2:\ Find\ the\ value\ of\ 2\ Sin\ 30^{0}\ Cos\ 30^{0}\ \hspace{15cm}$
$\color {black}{Solution:}\ 2\ Sin\ 30^{0}\ Cos\ 30^{0}\ = \hspace{18cm}$
$=\ Sin\ 2(30^{0})\ \hspace{15cm}$
$=\ Sin\ 60^{0}\ \hspace{15cm}$
$=\ \frac{\sqrt{3}}{2}\ \hspace{15cm}$
$Ex\ 3:\ Find\ the\ value\ of\ Cos^2\ 15^{0}\ -\ Sin^2\ 15^{0}\ \hspace{15cm}$
$\color {black}{Solution:}\ Cos^2\ 15^{0}\ -\ Sin^2\ 15^{0}\ =\ \hspace{18cm}$
$=\ Cos\ 2(15^{0})\ \hspace{15cm}$
$=\ Cos\ 30^{0}\ \hspace{15cm}$
$=\ \frac{\sqrt{3}}{2}\ \hspace{15cm}$
$Ex\ 4:\ Find\ the\ value\ of\ 1\ -\ 2\ Sin^2\ 45^{0}\ \hspace{15cm}$
$\color {black}{Solution:}\ 1\ -\ 2\ Sin^2\ 45^{0}\ =\ \hspace{18cm}$
$=\ Cos\ 2(45^{0})\ \hspace{15cm}$
$=\ Cos\ 90^{0}\ \hspace{15cm}$
$=\ 0\ \hspace{15cm}$
$Ex\ 5:\ Prove\ that\ Cos^4\ A\ -\ Sin^4\ A\ =\ Cos\ 2A\ \hspace{15cm}$
$\color {black}{Solution:}\ L.\ H.\ S\ =\ Cos^4\ A\ -\ Sin^4\ A\ \hspace{18cm}$
$=\ (Cos^2\ A)^2\ -\ (Sin^2\ A)^2\hspace{15cm}$
$=\ (Cos^2\ A\ +\ Sin^2\ A)\ (Cos^2\ A\ +\ Sin^2\ A)\hspace{10cm}$
$=\ (1)\ (Cos^2\ A\ -\ Sin^2\ A)\hspace{10cm}$
$=\ Cos\ 2A\ =\ R.H.S\ \hspace{15cm}$
$Ex\ 6:\ If\ Tan\ A\ =\ \frac{1}{3} \ and\ Tan\ B\ =\ \frac{1}{7},\ Show\ that\ 2A\ +\ B\ =\ \frac{π}{4}\ \hspace{15cm}$
$\color {black}{Solution:}\ Given\ Tan\ A\ =\ \frac{1}{3} \ and\ Tan\ B\ =\ \frac{1}{7}\ \hspace{18cm}$
$Tan(A + B)\ =\ \frac{Tan A\ +\ Tan B}{1\ -\ Tan A\ Tan B}\ \hspace{15cm}$
$Tan(2A + B)\ =\ \frac{Tan 2A\ +\ Tan B}{1\ -\ Tan 2A\ Tan B}\ —————- (1)\ \hspace{15cm}$
$Tan\ 2A\ =\ \frac{2\ Tan\ A}{1\ -\ Tan^2\ A}\ \hspace{10cm}$
$=\ \frac{2\ (\frac{1}{3})}{1\ -\ (\frac{1}{3})^2}\ \hspace{10cm}$
$=\ \frac{\frac{2}{3}}{1\ -\ \frac{1}{9}}\ \hspace{10cm}$
$=\ \frac{\frac{2}{3}}{\frac{8}{9}}\ \hspace{10cm}$
$=\ \frac{2}{3}\ ×\ \frac{8}{9}\ \hspace{10cm}$
$=\ \frac{3}{4}\ \hspace{10cm}$
$Equation\ (1)\ becomes$
$Tan(2A + B)\ =\ \frac{\frac{3}{4}\ +\ \frac{1}{7}}{1\ -\ \frac{3}{4}\ ×\ \frac{1}{7}}\ \hspace{15cm}$
$=\ \frac{\frac{21\ +\ 4}{28}}{1\ -\ \frac{3}{28}}\ \hspace{10cm}$
$=\ \frac{\frac{25}{28}}{\frac{28\ -\ 3}{28}}\ \hspace{10cm}$
$=\ \frac{\frac{25}{28}}{\frac{25}{28}}\ \hspace{10cm}$
$=\ 1\ \hspace{10cm}$
$Tan\ (2A\ +\ B)\ =\ 1\ \hspace{10cm}$
$\implies\ (2A\ +\ B)\ =\ Tan\ ^{-1}\ (1)\ \hspace{10cm}$
$\implies\ (2A\ +\ B)\ =\ \frac{π}{4}\ \hspace{10cm}$
$\color {royalblue} {Multiple\ Angles\ of\ 3A}:\ \hspace{20cm}$
$\color {brown} {Formulae}:\ \hspace{20cm}$
$1.\ Sin\ 3A\ =\ 3\ Sin\ A\ -\ 4\ Sin^3\ A\ \hspace{10cm}$
$2.\ Cos\ 3A\ =\ 4\ Cos^3\ A\ -\ 3\ Cos\ A\ \hspace{10cm}$
$3.\ Tan\ 3A\ =\ \frac{3\ Tan\ A\ -\ Tan^3\ A}{1\ -\ 3\ Tan^2\ A}\ \hspace{10cm}$
$Ex1:\ Find\ the\ value\ of\ 3\ Sin\ 10^{0}\ -\ 4\ Sin^3\ 10^{0}\ \hspace{15cm}$
$\color {black}{Solution:}\ 3\ Sin\ 10^{0}\ -\ 4\ Sin^3\ 10^{0}\ = \hspace{18cm}$
$=\ Sin\ 3(10^{0})\ \hspace{15cm}$
$=\ Sin\ 30^{0}\ \hspace{15cm}$
$=\ \frac{1}{2}\ \hspace{15cm}$
$Ex\ 2:\ If\ Sin\ θ\ =\ \frac{1}{3},\ find\ Sin\ 3θ\ \hspace{15cm}$
$\color {black}{Solution:}\ Given\ Sin\ θ\ =\ \frac{1}{3}\ \hspace{18cm}$
$Sin\ 3θ\ =\ 3\ Sin\ θ\ -\ 4\ Sin^3\ θ\ \hspace{10cm}$
$=\ 3( \frac{1}{3})\ -\ 4(\frac{1}{3})^3\ \hspace{10cm}$
$=\ 3( \frac{1}{3})\ -\ 4(\frac{1}{3})^3\ \hspace{10cm}$
$=\ 1\ -\ 4(\frac{1}{27})\ \hspace{10cm}$
$=\ 1\ -\ \frac{4}{27}\ \hspace{10cm}$
$=\ \frac{27\ -\ 4}{27}\ \hspace{10cm}$
$=\ \frac{23}{27}\ \hspace{10cm}$
$Ex\ 3:\ Find\ the\ value\ of\ 4\ Cos^3\ 15^{0}\ -\ 3\ Cos\ 15^{0}\ \hspace{15cm}$
$\color {black}{Solution:}\ 4\ Cos^3\ 15^{0}\ -\ 3\ Cos\ 15^{0}\ = \hspace{18cm}$
$=\ Cos\ 3(15^{0})\ \hspace{15cm}$
$=\ Cos\ 45^{0}\ \hspace{15cm}$
$=\ \frac{1}{\sqrt{2}}\ \hspace{15cm}$
$Ex\ 4:\ Find\ the\ value\ of\ \frac{3\ Tan\ {20}^0\ +\ Tan^3\ {20}^0}{1\ -\ 3\ Tan^2\ {20}^0}\ \hspace{15cm}$
$\color {black}{Solution:}\ \frac{3\ Tan\ {20}^0\ +\ Tan^3\ {20}^0}{1\ -\ 3\ Tan^2\ {20}^0}\ =\ \hspace{18cm}$
$=\ Tan\ 3(20^{0})\ \hspace{15cm}$
$=\ Tan\ 60^{0}\ \hspace{15cm}$
$=\ \sqrt{3}\ \hspace{15cm}$
$Ex\ 5:\ Prove\ that\ \frac{Cos\ 3θ}{Cos\ θ}\ +\ \frac{Sin\ 3θ}{Sin\ θ}\ =\ 4\ Cos\ 2θ\ \hspace{15cm}$
$\color {black}{Solution:}\ L.\ H.\ S\ =\ \frac{Cos\ 3θ}{Cos\ θ}\ +\ \frac{Sin\ 3θ}{Sin\ θ}\ \hspace{18cm}$
$=\ \frac{4\ Cos^3\ θ\ -\ 3\ Cos\ θ}{Cos\ θ}\ +\ \frac{3\ Sin\ θ\ -\ 4\ Sin^3\ θ}{Sin\ θ}\ \hspace{15cm}$
$=\ \frac{Cos\ θ(4\ Cos^2\ θ\ -\ 3)}{Cos\ θ}\ +\ \frac{Sin\ θ(3\ -\ 4\ Sin^2\ θ)}{Sin\ θ}\ \hspace{10cm}$
$=\ 4\ Cos^2\ θ\ -\ 3\ +\ 3\ -\ 4\ Sin^2\ θ\ \hspace{10cm}$
$=\ 4\ (Cos^2\ θ\ -\ Sin^2\ θ)\ \hspace{10cm}$
$=\ 4\ (Cos^2\ θ\ -\ Sin^2\ θ)\ \hspace{10cm}$
$=\ 4\ Cos\ 2θ\ \hspace{10cm}$
$Ex\ 6:\ Show\ that\ \frac{Sin\ 3A}{1\ +\ 2\ Cos\ 2A}\ =\ Sin\ A\ \hspace{15cm}$
$\color {black}{Solution:}\ W.\ K.\ T\ Sin\ 3A\ =\ 3\ Sin\ A\ -\ 4\ Sin^3\ A\ and\ 1\ -\ 2\ Sin^2A\ =\ Cos\ 2A\ \hspace{15cm}$
$L.\ H.\ S\ =\ \frac{Sin\ 3A}{1\ +\ 2\ Cos\ 2A}\ \hspace{15cm}$
$=\frac{3\ Sin\ A\ -\ 4\ Sin^3\ A}{1\ +\ 2\ (1\ -\ 2\ Sin^2A)}\ \hspace{10cm}$
$=\frac{Sin\ A(3\ -\ 4\ Sin^2\ A)}{1\ +\ 2\ -\ 4\ Sin^2A}\ \hspace{10cm}$
$=\frac{Sin\ A(3\ -\ 4\ Sin^2\ A)}{(3\ -\ 4\ Sin^2A)}\ \hspace{10cm}$
$=\ Sin\ A\ =\ R.H.S\ \hspace{10cm}$
$Ex\ 7:\ Prove\ that\ \frac{1\ -\ Cos\ 3A}{1\ -\ Cos\ A}\ =\ (1\ +\ 2\ Cos\ A)^2\ \hspace{15cm}$
$\color {black}{Solution:}\ L.\ H.\ S\ =\ \frac{1\ -\ Cos\ 3A}{1\ -\ Cos\ A}\ \hspace{18cm}$
$=\ \frac{1\ -\ (4\ Cos^3\ A\ -\ 3\ Cos\ A)}{1\ -\ Cos\ A}\ \hspace{10cm}$
$=\ \frac{1\ -\ 4\ Cos^3\ A\ +\ 3\ Cos\ A}{1\ -\ Cos\ A}\ \hspace{10cm}$
$=\ \frac{1\ -\ 4\ Cos^3\ A\ +\ 4\ Cos\ A\ -\ Cos\ A}{1\ -\ Cos\ A}\ \hspace{10cm}$
$=\ \frac{1\ -\ Cos\ A\ -\ 4\ Cos\ A(1\ -\ Cos^2\ A)}{1\ -\ Cos\ A}\ \hspace{10cm}$
$=\ \frac{1\ -\ Cos\ A\ -\ 4\ Cos\ A(1\ -\ Cos\ A)(1\ +\ Cos\ A)}{1\ -\ Cos\ A}\ \hspace{10cm}$
$=\ 1\ +\ 4\ Cos\ A(1\ +\ Cos\ A)\ \hspace{10cm}$
$=\ 1\ +\ 4\ Cos\ A\ +\ 4\ Cos^2\ A\ \hspace{10cm}$
$=\ (1\ +\ 2\ Cos\ A)^2\ \hspace{10cm}$