Principle of Mathematical Induction

Principle of Mathematical Induction

Question

If        sinα space plus space sinβ space equals space straight a
         space space cos space straight alpha space plus space cos space straight beta space equals space straight b comma

show that:

sin left parenthesis straight alpha plus straight beta right parenthesis space equals space fraction numerator 2 ab over denominator straight a squared plus straight b squared end fraction

Answer
sin space straight alpha space plus space sin space straight beta space space equals space straight a space space space space space space space space space rightwards double arrow space space space space space space space space space space space 2 space sin open parentheses fraction numerator straight alpha plus straight beta over denominator 2 end fraction close parentheses space cos open parentheses fraction numerator straight alpha minus straight beta over denominator 2 end fraction close parentheses space equals space straight a          ...(i)
cos space straight alpha space plus space cos space straight beta space equals space straight b space space space space space space space space space space rightwards double arrow space space space space space space 2 cos open parentheses fraction numerator straight alpha plus straight beta over denominator 2 end fraction close parentheses space cos open parentheses fraction numerator straight alpha minus straight beta over denominator 2 end fraction close parentheses space equals space straight b            ...(ii)
Dividing (i) and (ii), we get
fraction numerator 2 sin open parentheses begin display style fraction numerator straight alpha plus straight beta over denominator 2 end fraction end style close parentheses space cos open parentheses begin display style fraction numerator straight alpha minus straight beta over denominator 2 end fraction end style close parentheses over denominator 2 cos open parentheses begin display style fraction numerator straight alpha plus straight beta over denominator 2 end fraction end style close parentheses space cos open parentheses begin display style fraction numerator straight alpha minus straight beta over denominator 2 end fraction end style close parentheses end fraction space equals space straight a over straight b space space space space space space rightwards double arrow space space space space space space tan open parentheses fraction numerator straight alpha plus straight beta over denominator 2 end fraction close parentheses space equals space straight a over straight b

sin left parenthesis straight alpha plus straight beta right parenthesis space equals space fraction numerator 2 space tan open parentheses begin display style fraction numerator straight alpha plus straight beta over denominator 2 end fraction end style close parentheses over denominator 1 plus tan squared open parentheses begin display style fraction numerator straight alpha plus straight beta over denominator 2 end fraction end style close parentheses end fraction                 open parentheses because space sin space 2 straight x space equals space fraction numerator 2 space tanx over denominator 1 plus tan squared straight x end fraction close parentheses
               = fraction numerator 2 open parentheses begin display style straight a over straight b end style close parentheses over denominator 1 plus begin display style straight a squared over straight b squared end style end fraction space equals space fraction numerator 2 straight a over denominator straight b end fraction space cross times space fraction numerator straight b squared over denominator straight b squared plus straight a squared end fraction space equals space fraction numerator 2 ab over denominator straight a squared plus straight b squared end fraction

Hence,     sin left parenthesis straight alpha plus straight beta right parenthesis space equals space fraction numerator 2 ab over denominator straight a squared plus straight b squared end fraction

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