# 3.3 SUM AND PRODUCT FORMULAE & INVERSE TRIGONOMETRIC FUNCTIONS

$\color {brown} {Formulae}:\ \hspace{20cm}$
$1.\ Sin\ A\ +\ Sin\ B =\ 2\ Sin\ (\frac{A\ +\ B}{2})\ Cos\ (\frac{A\ -\ B}{2})\ \hspace{10cm}$
$2.\ Sin\ A\ -\ Sin\ B =\ 2\ Cos\ (\frac{A\ +\ B}{2})\ Sin\ (\frac{A\ -\ B}{2})\ \hspace{10cm}$
$3.\ Cos\ A\ +\ Cos\ B =\ 2\ Cos\ (\frac{A\ +\ B}{2})\ Cos\ (\frac{A\ -\ B}{2})\ \hspace{10cm}$
$4.\ Cos\ A\ -\ Cos\ B =\ -\ 2\ Sin\ (\frac{A\ +\ B}{2})\ Sin\ (\frac{A\ -\ B}{2})\ \hspace{10cm}$
$5.\ Sin\ A\ Cos\ B\ =\ \frac{1}{2}[Sin\ (A\ +\ B)\ +\ Sin\ (A\ -\ B)]\ \hspace{10cm}$
$6.\ Cos\ A\ Sin\ B\ =\ \frac{1}{2}[Sin\ (A\ +\ B)\ -\ Sin\ (A\ -\ B)]\ \hspace{10cm}$
$7.\ Cos\ A\ Cos\ B\ =\ \frac{1}{2}[Cos\ (A\ +\ B)\ +\ Cos\ (A\ -\ B)]\ \hspace{10cm}$
$8.\ Sin\ A\ Sin\ B\ =\ \frac{-\ 1}{2}[Cos\ (A\ +\ B)\ -\ Cos\ (A\ -\ B)]\ \hspace{10cm}$
$\color {royalblue} {Results}:\ \hspace{20cm}$
$1.\ Sin\ (A\ +\ B)Sin\ (A\ -\ B)\ =\ Sin^2\ A\ -\ Sin^2\ B\ \hspace{10cm}$
$2.\ Cos\ (A\ +\ B)Cos\ (A\ -\ B)\ =\ Cos^2\ A\ -\ Sin^2\ B\ \hspace{10cm}$
$\color {royalblue} {Examples}:\ \hspace{20cm}$
$Ex\ 1:\ Express\ the\ product\ 2\ Cos\ 5A\ Cos\ 3A\ as\ a\ sum\ \hspace{15cm}$
$\color {black}{Solution:}\ W.\ K.\ T.\ 2\ Cos\ A\ Cos\ B\ =\ Cos\ (A\ +\ B)\ +\ Cos\ (A\ -\ B)\ \hspace{18cm}$
$2\ Cos\ 5A\ Cos\ 3A\ =\ Cos\ (5A\ +\ 3A)\ +\ Cos\ (5A\ -\ 3A)\ \hspace{10cm}$
$=\ Cos\ 8A\ +\ Cos\ 2A\ \hspace{10cm}$
$Ex\ 2:\ Express\ Sin\ 5A\ -\ Sin\ 3A\ as\ product\ \hspace{15cm}$
$\color {black}{Solution:}\ W.\ K.\ T.\ Sin\ A\ -\ Sin\ B =\ 2\ Cos\ (\frac{A\ +\ B}{2})\ Sin\ (\frac{A\ -\ B}{2})\ \hspace{18cm}$
$Sin\ 5A\ -\ Sin\ 3A\ =\ 2\ Cos\ (\frac{5A\ +\ 3A}{2})\ Sin\ (\frac{5A\ -\ 3A}{2})\ \hspace{10cm}$
$=\ 2\ Cos\ (\frac{8A}{2})\ Sin\ (\frac{2A}{2})\ \hspace{10cm}$
$=\ 2\ Cos\ 4A\ Sin\ A\ \hspace{10cm}$
$Ex\ 3:\ Find\ the\ product\ value\ of\ Cos\ 70^{0}\ +\ Cos\ 10^{0}\ \hspace{15cm}$
$\color {black}{Solution:}\ W.\ K.\ T.\ Cos\ A\ +\ Cos\ B =\ 2\ Cos\ (\frac{A\ +\ B}{2})\ Cos\ (\frac{A\ -\ B}{2})\ \hspace{18cm}$
$Cos\ 70^{0}\ +\ Cos\ 10^{0} =\ 2\ Cos\ (\frac{70^{0}\ +\ 10^{0}}{2})\ Cos\ (\frac{70^{0}\ -\ 10^{0}}{2})\ \hspace{10cm}$
$=\ 2\ Cos\ (\frac{80^{0}}{2})\ Cos\ (\frac{60^{0}}{2})\ \hspace{10cm}$
$=\ 2\ Cos\ 40^{0}\ Cos\ 30^{0}\ \hspace{10cm}$
$=\ 2\ Cos\ 40^{0}\ \frac{\sqrt{3}}{2}\ \hspace{10cm}$
$=\ \sqrt{3}\ Cos\ 40^{0}\ \hspace{10cm}$
$Ex\ 4:\ Prove\ that\ \frac{Sin\ 3A\ +\ Sin\ A}{Cos\ 3A\ +\ Cos\ A}\ =\ Tan\ 2A\ \hspace{15cm}$
$\color {black}{Solution:}\ \hspace{18cm}$
$W.\ K.\ T\ Sin\ A\ +\ Sin\ B =\ 2\ Sin\ (\frac{A\ +\ B}{2})\ Cos\ (\frac{A\ -\ B}{2})\ and$
$Cos\ A\ +\ Cos\ B =\ 2\ Cos\ (\frac{A\ +\ B}{2})\ Cos\ (\frac{A\ -\ B}{2})$
$L.\ H.\ S\ =\ \frac{Sin\ 3A\ +\ Sin\ A}{Cos\ 3A\ +\ Cos\ A}\ \hspace{10cm}$
$=\ \frac{2\ Sin\ (\frac{3A\ +\ A}{2})\ Cos\ (\frac{3A\ -\ A}{2})}{2\ Cos\ (\frac{3A\ +\ A}{2})\ Cos\ (\frac{3A\ -\ A}{2})}\ \hspace{10cm}$
$=\ \frac{2\ Sin\ (\frac{4A}{2})\ Cos\ (\frac{2A}{2})}{2\ Cos\ (\frac{4A}{2})\ Cos\ (\frac{2A}{2})}\ \hspace{10cm}$
$=\ \frac{Sin\ 2A\ Cos\ A}{Cos\ 2A\ Cos\ A}\ \hspace{10cm}$
$=\ \frac{Sin\ 2A}{Cos\ 2A}\ \hspace{10cm}$
$=\ Tan\ 2A\ =\ R.H.S\ \hspace{10cm}$
$Ex\ 5:\ Prove\ that\ Sin\ 40^{0}\ +\ Sin\ 20^{0}\ -\ Cos\ 20^{0}\ =\ 0\ \hspace{15cm}$
$\color {black}{Solution:}\ W.\ K.\ T.\ Sin\ A\ +\ Sin\ B =\ 2\ Sin\ (\frac{A\ +\ B}{2})\ Cos\ (\frac{A\ -\ B}{2})\ \hspace{18cm}$
$L.\ H.\ S\ =\ Sin\ 40^{0}\ +\ Sin\ 20^{0}\ -\ Cos\ 20^{0}\ \hspace{10cm}$
$=\ 2\ Sin\ (\frac{40^{0}\ +\ 20^{0}}{2})\ Cos\ (\frac{40^{0}\ -\ 20^{0}}{2})\ -\ Cos\ 20^{0}\ \hspace{10cm}$
$=\ 2\ Sin\ 30^{0}\ Cos\ 10^{0}\ -\ Cos\ 10^{0}\ \hspace{10cm}$
$=\ 2\ ×\ \frac{1}{2}\ Cos\ 10^{0}\ -\ Cos\ 10^{0}\ \hspace{10cm}$
$=\ 0\ =\ R.H.S\ \hspace{10cm}$
$Ex\ 6:\ Prove\ that\ Sin\ 40^{0}\ +\ Sin\ 20^{0}\ =\ Sin\ 80^{0}\ \hspace{15cm}$
$\color {black}{Solution:}\ W.\ K.\ T.\ Sin\ A\ +\ Sin\ B =\ 2\ Sin\ (\frac{A\ +\ B}{2})\ Cos\ (\frac{A\ -\ B}{2})\ \hspace{18cm}$
$L.\ H.\ S\ =\ Sin\ 40^{0}\ +\ Sin\ 20^{0}\ \hspace{10cm}$
$=\ 2\ Sin\ (\frac{40^{0}\ +\ 20^{0}}{2})\ Cos\ (\frac{40^{0}\ -\ 20^{0}}{2})\ \hspace{10cm}$
$=\ 2\ ×\ \frac{1}{2}\ Cos\ 10^{0}\ \hspace{10cm}$
$=\ Cos\ 10^{0}\ \hspace{10cm}$
$=\ Sin\ (90^{0}\ -\ 10^{0})\ \hspace{3cm}\ \because\ Cos\ θ\ =\ Sin(90^{0}\ -\ θ)$
$=\ Sin\ 80^{0}\ =\ R.H.S\ \hspace{10cm}$
$Ex\ 7:\ Show\ that\ ( cos α + cos β )^2\ +\ ( sin α + sin β )^2\ =\ 4\ cos^2 (\frac{α- β}{2})\ \hspace{15cm}$
$\color {black}{Solution:}\ \hspace{19cm}$
$W.\ K.\ T\ Sin\ A\ +\ Sin\ B =\ 2\ Sin\ (\frac{A\ +\ B}{2})\ Cos\ (\frac{A\ -\ B}{2})\ and$
$Cos\ A\ +\ Cos\ B =\ 2\ Cos\ (\frac{A\ +\ B}{2})\ Cos\ (\frac{A\ -\ B}{2})$
$L.\ H.\ S\ =\ ( cos α + cos β )^2\ +\ ( sin α + sin β )^2\ \hspace{10cm}$
$=\ [2\ Cos\ (\frac{α\ +\ β}{2})\ Cos\ (\frac{α\ -\ β}{2})]^2\ +\ [2\ Sin\ (\frac{α\ +\ β}{2})\ Cos\ (\frac{α\ -\ β}{2})]^2\ \hspace{10cm}$
$=\ 4\ Cos^2\ (\frac{α\ +\ β}{2})\ Cos^2\ (\frac{α\ -\ β}{2})\ +\ 4\ Sin^2\ (\frac{α\ +\ β}{2})\ Cos^2\ (\frac{α\ -\ β}{2})\ \hspace{10cm}$
$=\ 4\ Cos^2\ (\frac{α\ -\ β}{2})[Cos^2\ (\frac{α\ +\ β}{2})\ +\ 4\ Sin^2\ (\frac{α\ +\ β}{2})]\ \hspace{10cm}$
$=\ 4\ Cos^2\ (\frac{α\ -\ β}{2})\ =\ R.H.S\ \hspace{10cm}$
$Ex\ 7:\ If\ a\ =\ Cos\ x\ +\ Cos\ y,\ b\ =\ Sin\ x\ -\ Sin\ y,\ Show\ that\ a^2\ +\ b^2\ =\ 4\ Cos^2\ (\frac{x\ +\ y}{2})\ \hspace{15cm}$
$\color {black}{Solution:}\ \hspace{19cm}$
$W.\ K.\ T\ Cos\ A\ +\ Cos\ B =\ 2\ Cos\ (\frac{A\ +\ B}{2})\ Cos\ (\frac{A\ -\ B}{2})\ and$
$Sin\ A\ -\ Sin\ B =\ 2\ Cos\ (\frac{A\ +\ B}{2})\ Sin\ (\frac{A\ -\ B}{2})$
$L.\ H.\ S\ =\ a^2\ +\ b^2\ \hspace{10cm}$
$=\ (Cos\ x\ +\ Cos\ y)^2\ +\ (Sin\ x\ -\ Sin\ y)^2\ \hspace{10cm}$
$=\ [2\ Cos\ (\frac{x\ +\ y}{2})\ Cos\ (\frac{x\ -\ y}{2})]^2\ +\ [2\ Cos\ (\frac{x\ +\ y}{2})\ Sin\ (\frac{x\ -\ y}{2})]^2\ \hspace{10cm}$
$=\ 4\ Cos^2\ (\frac{x\ +\ y}{2})\ Cos^2\ (\frac{x\ -\ y}{2})\ +\ 4\ Cos^2\ (\frac{x\ +\ y}{2})\ Sin^2\ (\frac{x\ -\ y}{2}) \hspace{10cm}$
$=\ 4\ Cos^2\ (\frac{x\ +\ y}{2})[Cos^2\ (\frac{x\ -\ y}{2})\ +\ Sin^2\ (\frac{x\ -\ y}{2})]\ \hspace{10cm}$
$=\ 4\ Cos^2\ (\frac{x\ +\ y}{2})\ \hspace{10cm}$
$Ex\ 6:\ Prove\ that\ Sin\ 20^{0}\ Sin\ 40^{0}\ Sin\ 80^{0}\ =\ \frac{\sqrt{3}}{8}\ \hspace{15cm}$
$\color {black}{Solution:}\ \hspace{19cm}$
$W.\ K.\ T\ Sin\ (A\ +\ B)Sin\ (A\ -\ B)\ =\ Sin^2\ A\ -\ Sin^2\ B\ and$
$Sin\ 3A\ =\ 3\ Sin\ A\ -\ 4\ Sin^3\ A$
$L.\ H.\ S\ =\ Sin\ 20^{0}\ Sin\ 40^{0}\ Sin\ 80^{0}\ \hspace{10cm}$
$=\ Sin\ 20^{0}\ Sin\ (60^{0}\ -\ 20^{0}) Sin\ ((60^{0}\ +\ 20^{0})\ \hspace{10cm}$
$=\ Sin\ 20^{0}\ (Sin^2\ 60^{0}\ -\ Sin^2\ 20^{0})\ \hspace{10cm}$
$=\ Sin\ 20^{0}\ ((\frac{\sqrt{3}}{2})^2\ -\ Sin^2\ 20^{0})\ \hspace{10cm}$
$=\ Sin\ 20^{0}\ (\frac{3}{4}\ -\ Sin^2\ 20^{0})\ \hspace{10cm}$
$=\ \frac{1}{4}\ Sin\ 20^{0}\ (3\ -\ 4\ Sin^2\ 20^{0})\ \hspace{10cm}$
$=\ \frac{1}{4}(3\ Sin\ 20^{0}\ -\ 4\ Sin^3\ 20^{0})\ \hspace{10cm}$
$=\ \frac{1}{4}\ Sin\ 3(20^{0})\ \hspace{10cm}$
$=\ \frac{1}{4}\ Sin\ 60^{0}\ \hspace{10cm}$
$=\ \frac{1}{4}\ (\frac{\sqrt{3}}{2})\ \hspace{10cm}$
$=\ \frac{\sqrt{3}}{8}\ =\ R.H.S\ \hspace{10cm}$
$Ex\ 7:\ Prove\ that\ Cos\ 20^{0}\ Cos\ 40^{0}\ Cos\ 80^{0}\ =\ \frac{1}{8}\ \hspace{15cm}$
$\color {black}{Solution:}\ \hspace{19cm}$
$W.\ K.\ T\ Cos\ (A\ +\ B)Cos\ (A\ -\ B)\ =\ Cos^2\ A\ -\ Sin^2\ B\ and$
$Cos\ 3A\ =\ 4\ Cos^3\ A\ -\ 3\ Cos\ A$
$L.\ H.\ S\ =\ Cos\ 20^{0}\ Cos\ 40^{0}\ Cos\ 80^{0}\ \hspace{10cm}$
$=\ Cos\ 20^{0}\ Cos\ (60^{0}\ -\ 20^{0}) Cos\ ((60^{0}\ +\ 20^{0})\ \hspace{10cm}$
$=\ Cos\ 20^{0}\ (Cos^2\ 60^{0}\ -\ Sin^2\ 20^{0})\ \hspace{10cm}$
$=\ Cos\ 20^{0}\ ((\frac{1}{2})^2\ -\ Sin^2\ 20^{0})\ \hspace{10cm}$
$=\ Cos\ 20^{0}\ (\frac{1}{4}\ -\ (1\ -\ Cos^2\ 20^{0})\ \hspace{10cm}$
$=\ Cos\ 20^{0}\ [\frac{1\ -\ 4\ +\ 4\ Cos^2\ 20^{0}}{4}]\ \hspace{10cm}$
$=\ \frac{1}{4}\ Cos\ 20^{0}\ [-\ 3\ +\ 4\ Cos^2\ 20^{0}]\ \hspace{10cm}$
$=\ \frac{1}{4}(4\ Cos^3\ 20^{0}\ -\ 3\ Cos\ 20^{0})\ \hspace{10cm}$
$=\ \frac{1}{4}\ Cos\ 3(20^{0})\ \hspace{10cm}$
$=\ \frac{1}{4}\ Cos\ 60^{0}\ \hspace{10cm}$
$=\ \frac{1}{4}\ (\frac{1}{2})\ \hspace{10cm}$
$=\ \frac{1}{8}\ =\ R.H.S\ \hspace{10cm}$

## INVERSE TRIGONOMETRIC FUNCTIONS

$\color {royalblue} {Definition}:\ \hspace{20cm}$
$A\ function\ which\ is\ the\ reverse\ process\ of\ a\ trigonometric\ function\ is\ called\ the\ inverse\ trigonometric\ function.$
$Ex:\ sin^{-1}\ x,\ cos^{-1}⁡\ x,\ tan^{-1}\ x$
$\color {royalblue} {Properties}:\ \hspace{20cm}$
$a)\ Sin^{-1}\ (Sin\ x)\ =\ x$
$b)\ Cos^{-1}\ (Cos\ x)\ =\ x$
$c)\ Tan^{-1}\ (Tan\ x)\ =\ x$
$d)\ Tan^{-1}\ x\ + Tan^{-1}\ y\ =\ Tan^{-1}\ (\frac{x\ +\ y}{1\ -\ x\ y})$
$e)\ Tan^{-1}\ x\ - Tan^{-1}\ y\ =\ Tan^{-1}\ (\frac{x\ -\ y}{1\ +\ x\ y})$
$Ex\ 1:\ What\ is\ the\ value\ of\ Sin^{-1}\ (\frac{1}{2})\ +\ Cos^{-1}\ (\frac{1}{2})\ \hspace{15cm}$
$\color {black}{Solution:}\ \hspace{19cm}$
$Sin^{-1}\ (\frac{1}{2})\ +\ Cos^{-1}\ (\frac{1}{2})\ =\ 30^{0}\ +\ 60^{0}\ \hspace{10cm}$
$=\ 90^{0}\ \hspace{10cm}$
$Ex\ 2:\ Prove\ that\ Sin^{-1}\ (2x\ \sqrt{1\ -\ x^2})\ =\ 2\ Sin^{-1}\ x\ \hspace{15cm}$
$\color {black}{Solution:}\ \hspace{19cm}$
$Put\ x\ =\ Sin\ θ,\ \implies\ θ\ =\ Sin^{-1}\ x\ \hspace{10cm}$
$L.\ H.\ S\ =\ Sin^{-1}\ (2x\ \sqrt{1\ -\ x^2})\ \hspace{10cm}$
$=\ Sin^{-1}\ (2\ Sin\ θ\ \sqrt{1\ -\ Sin^2\ θ})\ \hspace{10cm}$
$=\ Sin^{-1}\ (2\ Sin\ θ\ \sqrt{Cos^2\ θ})\ \hspace{10cm}$
$=\ Sin^{-1}\ (2\ Sin\ θ\ Cos\ θ)\ \hspace{10cm}$
$=\ Sin^{-1}\ (Sin\ 2θ)\ \hspace{10cm}$
$=\ 2θ\ \hspace{10cm}$
$=\ 2\ Sin^{-1}\ x\ =\ R.H.S\ \hspace{10cm}$
$Ex\ 3:\ Prove:\ Tan^{-1}\ (\frac{2x}{1\ -\ x^2})\ =\ 2\ Tan^{-1}\ x\ \hspace{15cm}$
$\color {black}{Solution:}\ \hspace{19cm}$
$Put\ x\ =\ Tan\ θ,\ \implies\ θ\ =\ Tan^{-1}\ x\ \hspace{10cm}$
$L.\ H.\ S\ =\ Tan^{-1}\ (\frac{2x}{1\ -\ x^2})\ \hspace{10cm}$
$=\ Tan^{-1}\ (\frac{2\ Tan\ θ}{1\ -\ Tan^2\ θ})\ \hspace{10cm}$
$=\ Tan^{-1}\ (Tan\ 2θ)\ \hspace{10cm}$
$=\ 2\ θ\ \hspace{10cm}$
$=\ 2\ Tan^{-1}\ x\ =\ =\ R.H.S\ \hspace{10cm}$
$Ex\ 4:\ Prove\ that\ Sin^{-1}\ (3x\ -\ 4x^3)\ =\ 3\ Sin^{-1}\ x\ \hspace{15cm}$
$\color {black}{Solution:}\ \hspace{19cm}$
$Put\ x\ =\ Sin\ θ,\ \implies\ θ\ =\ Sin^{-1}\ x\ \hspace{10cm}$
$L.\ H.\ S\ =\ Sin^{-1}\ (3x\ -\ 4x^3)\ \hspace{10cm}$
$\ =\ Sin^{-1}\ (3Sin\ θ\ -\ 4Sin^3\ θ)\ \hspace{10cm}$
$=\ Sin^{-1}\ (Sin\ 3θ)\ \hspace{10cm}$
$=\ 3\ θ\ \hspace{10cm}$
$=\ 3\ Sin^{-1}\ x\ =\ R.H.S\ \hspace{10cm}$
$Ex\ 5:\ Show\ that\ Tan^{-1}\ x\ + Tan^{-1}\ y\ =\ Tan^{-1}\ (\frac{x\ +\ y}{1\ -\ x\ y})\ \hspace{15cm}$
$\color {black}{Solution:}\ \hspace{19cm}$
$Let\ Tan^{-1}\ (x)\ =\ A,\ \implies\ x\ =\ Tan\ A\ \hspace{10cm}$
$Let\ Tan^{-1}\ (y)\ =\ B,\ \implies\ y\ =\ Tan\ B\ \hspace{10cm}$
$W.\ K.\ T\ Tan(A + B)\ =\ \frac{Tan A\ +\ Tan B}{1\ -\ Tan A\ Tan B}\ \hspace{10cm}$
$Tan(A + B)\ =\ \frac{x\ +\ y}{1\ -\ x\ y}\ \hspace{10cm}$
$A + B\ =\ Tan^{-1}(\frac{x\ +\ y}{1\ -\ x\ y})\ \hspace{10cm}$
$Tan^{-1}\ x\ + Tan^{-1}\ y\ =\ Tan^{-1}\ (\frac{x\ +\ y}{1\ -\ x\ y})\ \hspace{10cm}$
$Ex\ 6:\ Show\ that\ 2\ Tan^{-1}\ (\frac{1}{3})\ =\ Tan^{-1}\ (\frac{3}{4})\ \hspace{15cm}$
$\color {black}{Solution:}\ \hspace{19cm}$
$L.\ H.\ S\ =\ 2\ Tan^{-1}\ (\frac{1}{3})\ \hspace{10cm}$
$=\ Tan^{-1}\ (\frac{1}{3})\ +\ Tan^{-1}\ (\frac{1}{3})\ \hspace{10cm}$
$=\ Tan^{-1}\ (\frac{\frac{1}{3}\ +\ \frac{1}{3}}{1\ -\ \frac{1}{3}\ ×\ \frac{1}{3}})\ \hspace{2cm}\ \because\ Tan^{-1}\ x\ + Tan^{-1}\ y\ =\ Tan^{-1}\ (\frac{x\ +\ y}{1\ -\ x\ y})$
$=\ Tan^{-1}\ (\frac{\frac{2}{3}}{\frac{8}{9}})\ \hspace{10cm}$
$=\ Tan^{-1}\ (\frac{3}{4})\ =\ R.H.S\ \hspace{10cm}$