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17thKVS-JMO2014-Solution3

17thKVS-JMO2014-Solution3

  3Q) Let sin x + sin y = a and cos x + cos y = b, Show that tan\(\large \frac{x}{2}\) + tan\(\large \frac{y}{2}\) are two roots of the equation:

                              (a² + b² + 2b)t² - 4at + (a² + b² - 2b) = 0

 Solution:

  Given a = sin x + sin y   and b = cos x + cos y , using sin sum rule and cos sum rule we get ,

  a = 2 \(\sin \left ( \frac{x+y}{2} \right )\cos \left ( \frac{x-y}{2} \right )\)

 b = 2 \(\cos \left ( \frac{x+y}{2} \right )\cos \left ( \frac{x-y}{2} \right )\)

 Given equation :  

 (a² + b² + 2b)t² - 4at + (a² + b² - 2b) = 0 .

   Divide the equation by   a² + b² + 2b  we get

   t² - \( \large \frac{4a}{a^{2} + b^{2} + 2b } \) t + \(\large \frac{a^{2} + b^{2} - 2b}{a^{2} + b^{2} + 2b }\) = 0 .

    Sum of the roots  = \( \large \frac{4a}{a^{2} + b^{2} + 2b} \)

   Product of the roots = \(\large \frac{a^{2} + b^{2} -2b}{a^{2} + b^{2} +2b} \)

  Now to prove that sum of the roots = \( \large tan (\frac{x}{2}\)) + tan(\( \large \frac{y}{2}\)) 

    (OR)  Product of the roots = \( \large tan(\frac{x}{2}) * tan(\frac{y}{2}) \) .

 Let us find the value of  :

  a² + b² + 2b = 4 \( sin^{2} \left(  \large \frac{x+y}{2} \right ) \) \(cos^{2}(  \large \frac{x-y}{2} ) \) + 4 \( cos^{2} ( \large \frac{x+y}{2} ) \) \(cos^{2} ( \large \frac{x-y}{2} \) )  +  4 cos\( ( \large \frac{x+y}{2} )cos ( \large \frac{x-y}{2} ) \)

     = 4 \( \large \cos ^{2}\left ( \frac{x-y}{2} \right ) \left ( \sin ^{2} \left ( \frac{x+y}{2} \right) + \cos ^{2}\left ( \frac{x+y}{2} \right )\right )\) + 4 \(\cos \left ( \frac{x+y}{2} \right )\cos \left ( \frac{x-y}{2} \right )\)

 = 4 \(\large \cos \left ( \frac{x-y}{2} \right )\left ( \cos \left ( \frac{x-y}{2} \right )+\cos \left ( \frac{x+y}{2} \right ) \right )\)
 = 4 \(\large\cos \left ( \frac{x-y}{2} \right )\left [ 2\cos \left ( \frac{x}{2} \right )\cos \left ( \frac{y}{2} \right ) \right ]\)

  = 8 \(\large\cos \left ( \frac{x-y}{2} \right ) \cos \left ( \frac{x}{2} \right )\cos \left ( \frac{y}{2} \right )\)

  Sum of the roots =  \( \LARGE \frac{4a}{a^{2} + b^{2} + 2b}\)

  = \(\large\frac {4\left ( 2\sin \left ( \frac{x+y}{2} \right )\cos \left ( \frac{x-y}{2} \right ) \right )}{8 \cos \left ( \frac{x-y}{2} \right )\cos \left ( \frac{x}{2} \right )\cos \left ( \frac{y}{2} \right )}\)

 = \(\large \frac { \sin \left (\frac{x+y}{2} \right ) }{\cos \left ( \frac{x}{2} \right )\cos \left ( \frac{y}{2} \right )}\)

 = \(\large \frac { \sin \left (\frac{x}{2} \right )\cos \left ( \frac{y}{2} \right )+\cos \left ( \frac{x}{2} \right )\sin \left ( \frac{y}{2} \right ) }{\cos \left ( \frac{x}{2} \right )\cos \left ( \frac{y}{2} \right )}\)

 = \(\large \tan \left ( \frac{x}{2} \right )+\tan \left ( \frac{y}{2} \right )\)

 Therefore sum of the roots = \( \large \tan\left(\frac{x} {2} \right) + \tan \left (\frac {y} {2} \right \)

 Similarly we prove product of roots = tan(x/2)*tan(y/2).

 Hence tan(x/2) and tan(y/2) are roots of given equation .

  

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