COMPOUND ANGLES

COMPOUND ANGLES
1)    If  \(A,B,C,\alpha ,\beta ,\gamma .....\)are any angles then \(A + B + C + A + B - C,A - B + C,\)   are called compound Angles.
Theorem : For all \(A,B \in R:\cos (A + B) = \cos A\cos B - \sin A\sin B\)
Proof :    Let the termial sides of angles A and Bin standard position cut the circle in p and \(Q'\) respectively.  Then \(\angle COP = A\) and \(\angle COQ = B\) .
 Let \(\angle POQ = B\) and \(\angle COR = - B\) such that the terminal sides of Angles A+B and -B are in standard position and cut the circle in Q and R respectively.

\(\because \,\,\,\,\angle COR = B\) (numerically) We have \(\angle POR = A + B\)
    clearly \(\Delta COQ \equiv \Delta POR\,\,(numerically) \Rightarrow CQ = PR\)
by definition,     
    we have R=[cos(-B), sin (-B)]=(cosB, -sin B)
we have   \(\left. \begin{gathered} P = (\cos A,\sin A), \hfill \\ Q = [\cos (A + B).\sin (A + B)] \hfill \\ \end{gathered} \right|\) R=[cos(-B), sin (-B)]=(cosB, -sin B)
we have CQ=PR
CQ²=PR²
\( \Rightarrow {[\cos (A + B) - 1]^2} + {[\sin (A + B) - 0]^2} = {(\cos A - \cos B)^2} + {(\sin A + \sin B)^2}\)
On simplification     , we have
    cos(A+B) = cosA . cosB - sin A . sinB
by putting B = -B
    cos(A-B) = cosA . cosB + sinA . sinB
Note I :Sin (A +B)    = sin A cos B + cos A . sin B
Sin (A + B) =     cos [90⁰ - (A + B)]
                   =     cos [(90⁰ - A) - B)]
                   =    cos[ (90⁰ - A) cos B + sin (900 - A) . cos (-B)]
        Sin (A + B)    =    sin A . cos B + cos A . sin B
        Put B = -B
     \(\therefore\) sin (A - B) =    sin A . cos B - cos A . sin B
Note II :
    i) cos (A + B) + cos (A - B)    =  2cos A . cos B
    ii) cos (A + B) - cos (A - B)    =  -2sin A . sin B
    iii) sin (A + B) + sin (A - B)    =  2 sin A cos B
    iv) sin (A + B) - sin (A - B)    =  2 cos A. sin B
Examples :
1) Find  cos(45⁰-A) cos(45⁰-B) - sin(45⁰-A) sin(45⁰-B)
Solution :
    Let \({45^ \circ } - A = \alpha ,{45^ \circ } - B = \beta \)
    \(\cos \alpha .\cos \beta - \sin \alpha .\sin \beta = \cos (\alpha + \beta )\)    
    \(= \cos [({45^ \circ } - A) + ({45^ \circ } - B)]\)        
     \(= \cos [({90^ \circ } - (A + B)]\)       
      \(\cos ({45^ \circ } - A).\cos ({45^ \circ } - B) - \sin ({45^ \circ } - A)\sin ({45^ \circ } - B) = \sin (A + B)\)      
2)The value of  sin(n+1) A. sin(n+2)B + cos(n+1)A.cos(n+2)B
Solution :
    Let    \((n + 1)A = \alpha \,\,\,\,\,\,\,\,\,\,(n + 2)A = \beta \,\)  
    \(\therefore \,\,\,\,\,\sin \alpha .\sin \beta + \cos \alpha .\cos \beta = \cos (\beta - \alpha )\)
     \(= \cos ((n + 2)A - (n + 1)A) = \cos A\)   
    cos(n+1)A.cos(n+2)A +sin(n+1)A.sin(n+2)A=cos
3) The value of  \(\sin \theta + \sin ({120^ \circ } + \theta ) - \sin ({120^ \circ } - \theta )\)
Solution :
    We know
\(\therefore \sin ({120^ \circ } + \theta ) - \sin ({120^ \circ } - \theta ) = 2\cos ({120^ \circ }).\sin \theta \)
\( = 2\left( { - \frac{1}{2}} \right).\sin \theta = - \sin \theta \)
\(\therefore \sin \theta + ({120^ \circ } + \theta ) - \sin ({120^ \circ } - \theta ) = + \sin \theta - \sin \theta = 0\)
\(\sin \theta + \sin ({120^ \circ } + \theta ) - \sin ({120^ \circ } - \theta ) = 0\)  

COMPOUND ANGLES

COMPOUND ANGLES
Theorem 2 : If none of the angles A, B, A+B, and A-B are an odd multiples of \(\frac{\pi }{2}\)
i)\(\tan (A + B) = \frac{{\sin (A + B)}}{{\cos (A + B)}}\)
\(= \frac{{\sin A\cos B + \cos A.\sin B}}{{\cos A\cos B - \sin A.\sin B}}\)
=\(= \frac{{\frac{{\sin A.\cos B}}{{\cos A.\cos B}} + \frac{{\cos A.\sin B}}{{\cos A.\cos B}}}}{{\frac{{\cos A.\cos B}}{{\cos A.\cos B}} - \frac{{\sin A\cos B}}{{\cos A.\sin B}}}}\)
\(= \frac{{\tan A + \tan B}}{{1 - \tan A.\tan B}}\)
=\(\tan (A + B) = \frac{{\tan A + \tan B}}{{1 - \tan A.\tan B}}\)
ii)\(\tan (A - B) = \frac{{\tan A - \tan B}}{{1 + \tan A.\tan B}}\)
Put B = -B in (i)
\(\therefore \tan (A + ( - B)) = \frac{{\tan A + \tan ( - B)}}{{1 - \tan A.\tan ( - B)}}\)
\(\tan (A - B) = \frac{{\tan A - \tan B}}{{1 + \tan A.\tan B}}\)

COMPOUND ANGLES

COMPOUND ANGLES
Theorem 3 : If none of the angles A, B, A + B, A - B are an Integral multiple of  \(\pi\), then 
i) \(\cot (A + B) = \frac{{\cot A.\cot B - 1}}{{\cot B + \cot A}}\)           (ii) \(\cot (A - B) = \frac{{\cot A.\cot B + 1}}{{\cot B - \cot A}}\)
Proof : 
i) \(\cot (A + B) = \frac{{\cot A.\cot B - 1}}{{\cot B + \cot A}}\)         s
   \(\frac{1}{{\tan (A + B)}}\,\,\,\, = \,\,\,\,\,\frac{1}{{\left( {\frac{{\tan A + \tan B}}{{1 - \tan A\tan B}}} \right)}}\) 
     \(= \frac{{1 - \tan A.\tan B}}{{\tan A + \tan B}}\)       
Multiply NR and  Dr by cot A. cot B
\(\cot (A + B) = \frac{{\cot A.\cot B - 1}}{{\cot B + \cot A}}\)
ii)\(\cot (A + B) = \frac{{\cot A.\cot B - 1}}{{\cot B + \cot A}}\,\,\,\,\,\,\,\,\,\,(1)\)
 by putting B = -B in (1)
\(\cot (A + ( - B)) = \frac{{\cot A.\cot ( - B) - 1}}{{\cot ( - B) + \cot A}}\)
\(= \frac{{ - \cot A.\cot B - 1}}{{ - \cot B + \cot A}}\)
=\(\cot (A - B) = \frac{{\cot A.\cot B + 1}}{{\cot B - \cot A}}\)
Examples
1) If  \(\cot \beta = 2\tan (\alpha - \beta ),\)then show that \(\tan \alpha = 2\tan \beta + \cot \beta \)
Solution :
\(\cot \beta = 2\tan (\alpha - \beta ) = 2\left( {\frac{{\tan \alpha - \tan \beta }}{{1 + \tan \alpha .\tan \beta }}} \right)\)    
\(\Rightarrow \cot \beta [1 + \tan \alpha .\tan \beta ] = 2(\tan \alpha - \tan \beta )\)
\( \Rightarrow \cot \beta + \tan \alpha = 2\tan \alpha - 2\tan \beta \)    
\(\Rightarrow \tan \alpha = 2\tan \beta + \cot \beta \)        
2)Show that \(\tan {70^ \circ } - \tan {20^ \circ } = 2\tan {40^ \circ } + 4\tan {10^ \circ }\)
Solution :
\(\tan {50^ \circ } = \tan ({70^ \circ } - {20^ \circ }) = \frac{{\tan {{70}^ \circ } - \tan {{20}^ \circ }}}{{1 + \tan {{70}^ \circ }.\tan {{20}^ \circ }}}\)
\(\tan {50^ \circ } = \frac{{\tan {{70}^ \circ } - \tan {{20}^ \circ }}}{{1 + \tan {{70}^ \circ }.\cot {{20}^ \circ }}}\) \(\left[ \begin{gathered} \tan {70^ \circ } = \tan ({90^0} - {20^0}) = \cot {20^0} \hfill \\ \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\tan {20^0}.\,\,\cot {20^0} = 1 \hfill \\ \end{gathered} \right]\)
\(\tan {50^ \circ } = \frac{{\tan {{70}^ \circ } - \tan {{20}^ \circ }}}{2}\)
2 tan50⁰ = tan 70⁰ - tan20⁰          ............(1)
    again
\(\tan ({40^0} + {10^0}) = \frac{{\tan {{40}^ \circ } + \tan {{10}^ \circ }}}{{1 - \tan {{40}^ \circ }\tan {{10}^ \circ }}}\)
\(\tan {50^ \circ }(1 - \tan {40^ \circ }\tan {10^ \circ }) = \tan {40^ \circ } + \tan {10^ \circ }\)
\(\tan {50^ \circ } - \tan {10^ \circ } = \tan {40^ \circ } + \tan {10^ \circ }\)    
\( \Rightarrow \tan {50^ \circ } = 2\tan {10^ \circ } + \tan {40^ \circ }\)
\(\therefore \tan {70^ \circ } - \tan {20^ \circ } = 2\tan {50^ \circ }\)----(2)
 from (1,2)
\(\tan {70^ \circ } - \tan {20^ \circ } = 2\tan {40^ \circ } + 4\tan {10^ \circ }\)
note (1):\(\tan (\frac{\pi }{4} + A) = \frac{{\tan \frac{\pi }{4} + \tan A}}{{1 - \tan \frac{\pi }{4}\tan A}}\)

        \(\tan (\frac{\pi }{4} + A) = \frac{{1 + \tan A}}{{1 - \tan A}}\) =\( = \frac{{1 + \frac{{\sin A}}{{\cos A}}}}{{1 - \frac{{\sin A}}{{\cos A}}}}\)
\(\tan (\frac{\pi }{4} + A) = \frac{{\cos A + \sin A}}{{\cos A - \sin A}}\)
ii)Similarly,\(\tan (\frac{\pi }{4} - A) = \frac{{1 - \tan A}}{{1 + \tan A}} = \frac{{1 - \frac{{\sin A}}{{\cos A}}}}{{1 + \frac{{\sin A}}{{\cos A}}}}\)
\(\tan (\frac{\pi }{4} - A) = \frac{{1 - \tan A}}{{1 + \tan A}} = \frac{{\cos A - \sin A}}{{\cos A + \sin A}}\)

COMPOUND ANGLES

 COMPOUND ANGLES
Theorem 4:
Prove that : 1)sin(A+B) . sin(A-B) =sin²A -sin²B (or) cos²B - cos²A
2)cos(A+B).cos(A-B) =cos²A -sin²B (or) cos²B - sin²A
3)\(\tan (A + B).\tan (A - B) = \frac{{{{\tan }^2}A - {{\tan }^2}B}}{{1 - {{\tan }^2}A{{\tan }^2}B}}\)    
4) \(cot(A + B).cot(A - B) = \frac{{co{t^2}A\,\,\,.\,\,co{t^2}B - 1}}{{co{t^2}B - co{t^2}A}}\)   
Proof : sin(A+B) . sin(A-B)
      \(= [\sin A.\cos B + \cos A.\sin B][\sin A.\cos B - \cos A.\sin B]\) ((a+b)(a-b)= (a²-b²))    
     = \({\sin ^2}A.{\cos ^2}B - {\cos ^2}A.{\sin ^2}B\)
     = \({\sin ^2}A(1 - {\sin ^2}B) - (1 - {\sin ^2}A).{\sin ^2}B\)
     =\(= {\sin ^2}A - {\sin ^2}A.{\sin ^2}B - {\sin ^2}B + {\sin ^2}A.{\sin ^2}B\)
      \(= {\sin ^2}A - {\sin ^2}B. = 1 - {\cos ^2}A - 1 + {\cos ^2}B = {\cos ^2}B - {\cos ^2}A\)
     \(\therefore \sin (A + B).\sin (A - B) = {\sin ^2}A - {\sin ^2}B = {\cos ^2}B - {\cos ^2}A\)
Similarly We can prove that (2), (3) & (4)

COMPOUND ANGLES

 COMPOUND ANGLES
Theorem 5 : 
Prove that for any three non-zero Angles A, B, C
i)\(\sin (A + B + C) = \sin A.\cos B.\cos C + \cos A.\sin B.\cos C\) \(+ \cos A.\cos B.\sin C - \sin A.\sin B.\sin C\)
=\(\sum \)sinA  cosB cosC -sinA
ii)\(\cos (A + B + C) = \cos A.\cos B.\cos C - \sin A.\sin B.\cos C\) \(- \sin A.\cos B.\sin C - \sin A.\cos B.\sin C\)
iii)\(\left. \begin{gathered} tan(A + B + C) = \frac{{{s_1} - {s_3}}}{{1 - {s_2}}} \hfill \\ \cot (A + B + C) = \frac{{{s_1} - {s_3}}}{{1 - {s_2}}} \hfill \\ \end{gathered} \right\}\begin{array}{*{20}{c}} {{s_1} = \Sigma \tan A\,\,(or)\,\,\Sigma \cot A} \\ {{s_2} = \Sigma \tan A\,\tan B\,(or)\,\,\Sigma \cot A} \\ {{s_3} = \pi \tan A\,\,(or)\,\,\pi \cot A} \end{array}\cot B\)
Proof :
 i)   \(\sin (A + B + C) = \sin ((A + B) + C)\) 
      \(= \sin (A + B)\cos C + \cos (A + B)\sin C\)        
      \(= (\sin A.\cos B + \cos A.\sin B)\cos C\)    
  sin(A+B+C) = sinA .cosB cosC +cosA . sinB . cosC +cosA. cosB. sinC -sinA.sinB sinC
ii)\(\cos (A + B + C) = \cos (A + B)\cos C - \sin (A + B).\sin C\)
  \(= (\cos A.\cos B - \sin A.\sin B)\cos C - (\sin A.\cos B + \cos A.\sin B)\sin C\)                 
cos(A+B+C) = cosA . cosB.cosC - sinA. sinB.cosC - sinA.cosB sinC-sinA.cosB.sinC
iii) \(\tan ((A + B) + C) = \frac{{\tan (A + B) + \tan C}}{{1 - \tan (A + B).\tan C}}\)
\(= \frac{{\left( {\frac{{\tan A + \tan B}}{{1 - \tan A.\tan B}}} \right) + \tan C}}{{1 - \left( {\frac{{\tan A + \tan B}}{{1 - \tan A.\tan B}}} \right).\tan C}}\)
\(= \frac{{(\tan A + \tan B + \tan C) - (\tan A.TanB.\tan C)}}{{1 - [\tan A.\tan B + \tan B.\tan C + \tan C.\tan A}}\)
\(\tan (A + B + C) = \frac{{{S_1} - {S_3}}}{{1 - {S_2}}}\)
iv)    Similarly, We can prove that 
\(\cot (A + B + C) = \frac{{(\cot A + \cot B + \cot C) - (\cot A.\cot B.\cot C)}}{{1 - [\cot A.\cot B + \cot B.\cot C + \cot C.\cot A]}}\)
\(\cot (A + B + C) = \frac{{{S_1} - {S_3}}}{{1 - {S_2}}}\)
Find the values \(\sin {15^ \circ },\cos {15^ \circ },\tan {15^ \circ },\tan {75^ \circ },\cos {75^ \circ },\tan {75^ \circ }\) and so on.
Solution:
\(\sin {15^ \circ } = \sin ({45^ \circ } - {30^ \circ })\)
\(= \sin {45^ \circ }.\cos {30^ \circ } - \cos {45^ \circ }.\sin {30^ \circ }\)
\( = \frac{1}{{\sqrt 2 }}.\frac{{\sqrt 3 }}{2} - \frac{1}{{\sqrt 2 }}.\frac{1}{2}\)
\(\sin {15^ \circ } = \frac{{\sqrt 3 - 1}}{{2\sqrt 2 }}\)
cos15⁰ = cos(45⁰ -30⁰)     
\(= \cos {45^ \circ }.\cos {30^ \circ } + \sin {45^ \circ }.\sin {30^ \circ }\)     
 \(= \frac{1}{{\sqrt 2 }}.\frac{{\sqrt 3 }}{2} + \frac{1}{{\sqrt 2 }}.\frac{1}{2}\)    
 \(\cos {15^ \circ } = \frac{{\sqrt 3 + 1}}{{2\sqrt 2 }}\)
 tan15⁰ = tan(45⁰ -30⁰)  
\(= \frac{{\tan {{45}^ \circ } - \tan {{30}^ \circ }}}{{1 + \tan {{45}^ \circ }.\tan {{30}^ \circ }}}\)
\(= \frac{{1 - \frac{1}{{\sqrt 3 }}}}{{1 - \frac{1}{{\sqrt 3 }}}}\)
\(= \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} = \frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)}}\)
\( = \frac{{2(2 - \sqrt {3)} }}{2}\)
tan15⁰ = 2-\(\sqrt{3}\)
Note : 
\(\csc {15^ \circ } = \frac{{2\sqrt 2 }}{{\sqrt 3 - 1}} = \frac{{2\sqrt 2 (\sqrt 3 + 1)}}{2} = \sqrt 2 (\sqrt 3 + 1)\)        
\(\sec {15^ \circ } = \frac{{2\sqrt 2 }}{{\sqrt 3 + 1}} = \frac{{2\sqrt 2 (\sqrt 3 - 1)}}{2} = \sqrt 2 (\sqrt 3 - 1)\)
\(\cot {15^ \circ } = \frac{1}{{\tan {{15}^ \circ }}} = \frac{1}{{2 - \sqrt 3 }} = \frac{{2 + \sqrt 3 }}{{\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}} = 2 + \sqrt 3 \)        
\(\sin {75^o} = \sin ({90^ \circ } - {15^ \circ }) = \cos {15^ \circ } = \frac{{\sqrt 3 + 1}}{{2\sqrt 2 }}\)     
\(\cos {75^o} = \cos ({90^ \circ } - {15^ \circ }) = \sin {15^ \circ } = \frac{{\sqrt 3 - 1}}{{2\sqrt 2 }}\)   
\(\cot {75^o} = \cot ({90^ \circ } - {15^ \circ }) = \tan {15^ \circ } = 2 - \sqrt 3 \)
\(\csc {75^o} = \csc ({90^ \circ } - {15^ \circ }) = \sec {15^ \circ } = \sqrt 2 (\sqrt 3 - 1)\)
\(\sec {75^o} = \sec ({90^ \circ } - {15^ \circ }) = \csc {15^ \circ } = \sqrt 2 (\sqrt 3 + 1)\)
\(\tan {75^o} = \tan ({90^ \circ } - {15^ \circ }) = \cot {15^ \circ } = 2 + \sqrt 3 \)