Unit – III – 3.1 TRIGONOMETRY

\[\LARGE{\color {red}{CHAPTER\ 3.1:\ COMPOUND\ ANGLES\ (Text)}}\]
\[\color {royalblue} {Introduction}:\ \hspace{20cm}\]
\[Trigonometry\ is\ one\ of\ the\ oldest\ branches\ of\ Mathematics.\ The\ word\]
\[Trigonometry\ is\ derived\ from\ the\ Greek\ words\ ‘Trigonon’\ and\ ‘metron’\ means\]
\[measurement\ of\ angles.\ In\ olden\ days\ Trigonometry\ was\ mainly\ used\ as\ a\ tool\ for\]
\[studying\ astronomy.\ In\ earlier\ stages\ Trigonometry\ was\ mainly\ concerned\ with\ angles\]
\[of\ a\ triangle.\ But\ now\ it\ has \ its\ applications\ in\ various\ branches\ of\ science\ such\ as\]
\[surveying,\ engineering,\ navigations\ etc.\ For\ the\ study\ of\ higher\ mathematics,\ knowledge\ of\ Trigonometry\ is\ essential.\]
\[\color {royalblue} {Trigonometrical\ ratios}:\ \hspace{20cm}\]
\[There\ are\ six\ Trigonometrical\ ratios\ sine,\ cosine,\ tangent,\ cotangent,\ secant\ and\]\[cosecant\ shortly\ written\ as\ sin θ,\ cos θ,\ tan θ,\ cot θ,\ sec θ\ and\ cosec θ.\]
\[\color {royalblue} {Fundamental\ Trigonometrical\ identities}:\ \hspace{20cm}\]
\[1\ .\ sin^2θ\ +\ cos^2θ\ =\ 1\ \hspace{10cm}\]
\[2\ .\ 1\ +\ tan^2θ\ =\ sec^2θ\ \hspace{10cm}\]
\[3\ .\ 1\ +\ cot^2θ\ =\ cosec^2θ\ \hspace{10cm}\]
\[\color {royalblue} {Trigonometrical\ ratios\ of\ known\ angles}:\ \hspace{20cm}\]
CBSE Class 10 - Trigonometric Ratios of Some Specific Angles Offered by  Unacademy
\[\color {royalblue} {Signs\ of\ Trigonometrical\ ratios}:\ \hspace{20cm}\]
Trigonometry Quadrant Formulas
Quadrant  Sign of ratiosRemember
IAll are +ve   ( 90 –  θ,   360 +  θ )All
IISin  and cosec  are +ve, Other ratios are –ve   ( 90 +  θ,   180 –  θ )Silver  
IIItan θ and  cot θ  are +ve, Other ratios are –ve ( 180 +  θ,   270 –  θ )Tea
IVcos θ and  sec θ  are +ve, Other ratios are –ve ( 180 +  θ,   360 –  θ )Cups
\[\color {brown} {Working\ rule\ for\ 180 ± θ\ and\ 360 ± θ}:\ \hspace{20cm}\]
RatioFalls in QuadrantChange of ratio
Sin (180 +  θ )III-Sin θ
Sin (180 –  θ )IISin θ
Cos (180 +  θ )III– Cos θ
Cos (180 –  θ )II– Cos θ
Tan (180 +  θ )IIITan θ
Tan (180 –  θ )II– Tan θ
Sin ( 360 – θ )IV– Sin θ
\[\color {brown} {Working\ rule\ for\ 90 ± θ\}:\ \hspace{20cm}\]
RatioFalls in QuadrantChange of ratio
Sin (90 +  θ )IICos θ
Sin (90 –  θ )ICos θ
cos (90 +  θ )II-Sin  θ
Cos (90 –  θ )ICos θ
\[\color {royalblue} {Formulae}:\ \hspace{20cm}\]
\[1)\ Sin ( A + B )\ =\ Sin A\ Cos B\ +\ Cos A\ Sin B\]
\[2)\ Sin ( A – B )\ =\ Sin A\ Cos B\ -\ Cos A\ Sin B\]
\[3)\ Cos( A + B )\ =\ Cos A\ Cos B\ -\ Sin A Sin B\]
\[4)\ Cos( A – B )\ =\ Cos A\ Cos B\ +\ Sin A Sin B\]
\[5)\ Tan(A + B)\ =\ \frac{Tan A\ +\ Tan B}{1\ -\ Tan A\ Tan B}\]
\[6)\ Tan(A – B)\ =\ \frac{Tan A\ -\ Tan B}{1\ +\ Tan A\ Tan B}\]
\[\color {brown} {Note}:\ Sin\ ( – θ )\ =\ -\ sin\ θ\ \hspace{2cm}\ cos\ ( – θ )\ =\ cos\ θ\]
\[\color {royalblue} {Examples}:\ \hspace{20cm}\]
\[Ex1:\ Find\ the\ value\ of\ sin\ {50}^0\ cos\ {40}^0\ +\ cos\ {50}^0\ sin\ {40}^0\ \hspace{15cm}\]
\[\color {black}{Solution:}\ W.\ K.\ T\ Sin ( A + B )\ =\ Sin A\ Cos B\ +\ Cos A\ Sin B\ \hspace{18cm}\]
\[sin\ {50}^0\ cos\ {40}^0\ +\ cos\ {50}^0\ sin\ {40}^0\ =\ Sin\ ({50}^0\ + {40}^0)\ \hspace{10cm}\]
\[=\ Sin \ {90}^0\ \hspace{10cm}\]
\[=\ 1\ \hspace{10cm}\]
\[Ex2:\ Find\ the\ value\ of\ cos\ {40}^0\ cos\ {20}^0\ -\ sin\ {40}^0\ sin\ {20}^0\ \hspace{15cm}\]
\[\color {black}{Solution:}\ W.\ K.\ T\ Cos A\ Cos B\ -\ Sin A Sin B\ =\ Cos( A + B )\ \hspace{18cm}\]
\[cos\ {40}^0\ cos\ {20}^0\ -\ sin\ {40}^0\ sin\ {20}^0\ =\ cos\ ({40}^0\ + {20}^0)\ \hspace{10cm}\]
\[=\ Sin \ {60}^0\ \hspace{10cm}\]
\[=\ \frac{1}{2}\ \hspace{10cm}\]
\[Ex3:\ Find\ the\ value\ of\ \frac{Tan\ {20}^0\ +\ Tan\ {25}^0}{1\ -\ Tan\ {20}^0\ Tan\ {25}^0}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ W.\ K.\ T\ \frac{Tan A\ +\ Tan B}{1\ -\ Tan A\ Tan B}\ =\ Tan(A + B)\ \hspace{18cm}\]
\[\frac{Tan\ {20}^0\ +\ Tan\ {25}^0}{1\ -\ Tan\ {20}^0\ Tan\ {25}^0}\ =\ Tan(A + B)\ \hspace{10cm}\]
\[Ex4:\ If\ Tan\ A\ =\ \frac{10}{11} \ and\ Tan\ B\ =\ \frac{1}{21},\ Show\ that\ A\ +\ B\ =\ {45}^0\ \hspace{15cm}\]
\[\color {black}{Solution:}\ Given\ Tan\ A\ =\ \frac{10}{11} \ and\ Tan\ B\ =\ \frac{1}{21}\ \hspace{18cm}\]
\[W.\ K.\ T\ Tan(A + B)\ =\ \frac{Tan A\ +\ Tan B}{1\ -\ Tan A\ Tan B}\ \hspace{15cm}\]
\[=\ \frac{\frac{10}{11}\ +\ \frac{1}{21}}{1\ -\ \frac{10}{11}\ ×\ \frac{1}{21}}\ \hspace{10cm}\]
\[=\ \frac{\frac{210\ +\ 11}{231}}{1\ -\ \frac{10}{231}}\ \hspace{10cm}\]
\[=\ \frac{\frac{221}{231}}{\frac{231\ -\ 10}{231}}\ \hspace{10cm}\]
\[=\ \frac{\frac{221}{231}}{\frac{221}{231}}\ \hspace{10cm}\]
\[=\ 1\ \hspace{10cm}\]
\[Tan\ (A\ +\ B)\ =\ 1\ \hspace{10cm}\]
\[\implies\ (A\ +\ B)\ =\ Tan\ ^{-1}\ (1)\ \hspace{10cm}\]
\[\implies\ (A\ +\ B)\ =\ {45}^0\ \hspace{10cm}\]
\[Ex5:\ If\ Sin\ A\ =\ \frac{3}{5} \ and\ Cos\ B\ =\ \frac{12}{13},\ Prove\ that\ Sin(A\ +\ B)\ =\ \frac{56}{65}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ Given\ Sin\ A\ =\ \frac{3}{5} \ and\ Cos\ B\ =\ \frac{12}{13}\ \hspace{18cm}\]
\[W.\ K.\ T\ Sin ( A + B )\ =\ Sin A\ Cos B\ +\ Cos A\ Sin B\ \hspace{15cm}\]
\[Cos\ A\ =\ ?\ ,\ Sin\ B\ = ?\ \hspace{10cm}\]
\[Cos\ A\ =\ \sqrt{1\ -\ Sin^2\ A}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ (\frac{3}{5})^2}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ \frac{9}{25}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{25\ -\ 9}{25}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{16}{25}}\ \hspace{10cm}\]
\[Cos\ A\ =\ \frac{4}{5}\ \hspace{10cm}\]
\[Sin\ B\ =\ \sqrt{1\ -\ Sin^2\ B}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ (\frac{12}{13})^2}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ \frac{144}{169}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{169\ -\ 144}{169}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{25}{169}}\ \hspace{10cm}\]
\[Sin\ B\ =\ \frac{5}{13}\ \hspace{10cm}\]
\[Sin ( A + B )\ =\ (\frac{3}{5})\ (\frac{12}{13})\ +\ (\frac{4}{5})\ (\frac{5}{13})\ \hspace{10cm}\]
\[=\ \frac{36}{25}\ +\ \frac{20}{65}\ \hspace{10cm}\]
\[=\ \frac{36\ +\ 20}{65}\ \hspace{10cm}\]
\[=\ \frac{56}{65}\ \hspace{10cm}\]
\[Sin ( A + B )\ =\ \frac{56}{65}\ \hspace{10cm}\]
\[Ex6:\ If\ A\ and\ B\ are\ acute\ angles\ and\ Sin\ A\ =\ \frac{1}{\sqrt{10}},\ Sin\ B\ =\ \frac{1}{\sqrt{5}},\ \hspace{15cm}\]\[ Prove\ that\ (A\ +\ B)\ =\ \frac{π}{4}\ \hspace{10cm}\]
\[\color {black}{Solution:}\ Given\ Sin\ A\ =\ \frac{1}{\sqrt{10}} \ and\ Sin\ B\ =\ \frac{1}{\sqrt{5}}\ \hspace{18cm}\]
\[W.\ K.\ T\ Sin ( A + B )\ =\ Sin A\ Cos B\ +\ Cos A\ Sin B\ \hspace{15cm}\]
\[Cos\ A\ =\ ?\ ,\ Cos\ B\ = ?\ \hspace{10cm}\]
\[Cos\ A\ =\ \sqrt{1\ -\ Sin^2\ A}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ (\frac{1}{\sqrt{10}})^2}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ \frac{1}{10}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{10\ -\ 1}{10}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{9}{10}}\ \hspace{10cm}\]
\[Cos\ A\ =\ \frac{3}{\sqrt{10}}\ \hspace{10cm}\]
\[Cos\ B\ =\ \sqrt{1\ -\ Sin^2\ B }\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ (\frac{1}{\sqrt{5}})^2}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ \frac{1}{5}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{5\ -\ 1}{10}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{4}{5}}\ \hspace{10cm}\]
\[Cos\ B\ =\ \frac{2}{\sqrt{5}}\ \hspace{10cm}\]
\[Sin ( A + B )\ =\ (\frac{1}{\sqrt{10}})\ (\frac{2}{\sqrt{5}})\ +\ (\frac{3}{\sqrt{10}})\ (\frac{1}{\sqrt{5}})\ \hspace{10cm}\]
\[=\ \frac{2}{\sqrt{50}}\ +\ \frac{3}{\sqrt{50}}\ \hspace{10cm}\]
\[=\ \frac{2\ +\ 3}{\sqrt{50}}\ \hspace{10cm}\]
\[=\ \frac{5}{\sqrt{50}}\ \hspace{10cm}\]
\[=\ \frac{5}{\sqrt{25 × 2}}\ \hspace{10cm}\]
\[Sin ( A + B )\ =\ \frac{1}{\sqrt{2}}\ \hspace{10cm}\]
\[(A\ +\ B)\ =\ \frac{π}{4}\ \hspace{10cm}\]
\[Ex7:\ If\ Sin\ A\ =\ \frac{3}{5} \ and\ Cos\ B\ =\ \frac{12}{13},\ find\ the\ value\ of\ Cos(A\ +\ B)\ \hspace{15cm}\]
\[\color {black}{Solution:}\ Given\ Sin\ A\ =\ \frac{3}{5} \ and\ Cos\ B\ =\ \frac{12}{13}\ \hspace{18cm}\]
\[W.\ K.\ T\ Cos( A + B )\ =\ Cos A\ Cos B\ -\ Sin A\ Sin B\ \hspace{15cm}\]
\[Cos\ A\ =\ ?\ ,\ Sin\ B\ = ?\ \hspace{10cm}\]
\[Cos\ A\ =\ \sqrt{1\ -\ Sin^2\ A}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ (\frac{3}{5})^2}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ \frac{9}{25}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{25\ -\ 9}{25}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{16}{25}}\ \hspace{10cm}\]
\[Cos\ A\ =\ \frac{4}{5}\ \hspace{10cm}\]
\[Sin\ B\ =\ \sqrt{1\ -\ Sin^2\ B}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ (\frac{12}{13})^2}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ \frac{144}{169}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{169\ -\ 144}{169}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{25}{169}}\ \hspace{10cm}\]
\[Sin\ B\ =\ \frac{5}{13}\ \hspace{10cm}\]
\[Cos( A + B )\ =\ (\frac{4}{5})\ (\frac{12}{13})\ -\ (\frac{3}{5})\ (\frac{5}{13})\ \hspace{10cm}\]
\[=\ \frac{48}{65}\ -\ \frac{15}{65}\ \hspace{10cm}\]
\[=\ \frac{48\ -\ 15}{65}\ \hspace{10cm}\]
\[=\ \frac{33}{65}\ \hspace{10cm}\]
\[Cos ( A + B )\ =\ \frac{33}{65}\ \hspace{10cm}\]
\[Ex8:\ If\ A\ +\ B\ =\ {45}^0 \ Prove\ that\ (1\ +\ Tan\ A)\ (1\ +\ Tan\ B)\ =\ 2.\ \hspace{15cm}\]\[Hence\ deduce\ the\ value\ of\ Tan\ \ 22\ {\frac{1}{2}}^0\ \hspace{13cm}\]
\[\color {black}{Solution:}\ Given\ A\ +\ B\ =\ 45^{0}\ \hspace{18cm}\]
\[Taking\ Tan\ on\ both\ sides\ \hspace{10cm}\]
\[Tan\ (A\ +\ B)\ =\ Tan\ 45^{0}\ \hspace{10cm}\]
\[\frac{Tan A\ +\ Tan B}{1\ -\ Tan A\ Tan B}\ =\ 1\ \hspace{10cm}\]
\[Tan\ A\ +\ Tan\ B\ =\ 1\ -\ Tan\ A\ Tan\ B\ \hspace{10cm}\]
\[Tan\ A\ +\ Tan\ B\ +\ Tan\ A\ Tan\ B\ =\ 1\ ————\ (1)\ \hspace{10cm}\]
\[L.\ H.\ S\ =\ (1\ +\ Tan\ A)\ (1\ +\ Tan\ B)\ \hspace{10cm}\]
\[=\ 1\ +\ Tan\ B\ +\ Tan\ A\ +\ Tan\ A\ Tan\ B\ \hspace{10cm}\]
\[=\ 1\ +\ 1\ \hspace{5cm}\ using\ (1)\]
\[=\ 2\ =\ R.\ H.\ S\ \hspace{10cm}\]
\[(1\ +\ Tan\ A)\ (1\ +\ Tan\ B)\ =\ 2\ \hspace{10cm}\]
\[Put\ B\ =\ A\ \hspace{10cm}\]
\[A\ +\ A\ =\ 45^{0}\ \hspace{10cm}\]
\[2A\ =\ 45^{0}\ \hspace{10cm}\]
\[A\ =\ 22\ {\frac{1}{2}}^0\ \hspace{10cm}\]
\[(1\ +\ Tan\ 22\ {\frac{1}{2}}^0)\ (1\ +\ Tan\ 22\ {\frac{1}{2}}^0)\ =\ 2\ \hspace{10cm}\]
\[(1\ +\ Tan\ 22\ {\frac{1}{2}}^0)^2\ =\ 2\ \hspace{10cm}\]
\[1\ +\ Tan\ 22\ {\frac{1}{2}}^0\ =\ \sqrt{2}\ \hspace{10cm}\]
\[Tan\ 22\ {\frac{1}{2}}^0\ =\ \sqrt{2}\ -\ 1\hspace{10cm}\]