Principle Of Mathematical Induction

Question
CBSEENMA11014318

Prove the following:

sin space straight C space minus space sin space straight D space equals space 2 cos fraction numerator straight C plus straight D over denominator 2 end fraction sin fraction numerator straight C minus straight D over denominator 2 end fraction


Solution

Let fraction numerator straight C plus straight D over denominator 2 end fraction equals straight x space space and space fraction numerator straight C minus straight D over denominator 2 end fraction equals straight y
R.H.S. = 2 cos open parentheses fraction numerator straight C plus straight D over denominator 2 end fraction close parentheses space sin open parentheses fraction numerator straight C minus straight D over denominator 2 end fraction close parentheses space equals space 2 space cosx space siny
          = sin (x + y) - sin (x - y)
          = sin open parentheses fraction numerator straight C plus straight D over denominator 2 end fraction plus fraction numerator straight C minus straight D over denominator 2 end fraction close parentheses minus sin open parentheses fraction numerator straight C plus straight D over denominator 2 end fraction minus fraction numerator straight C minus straight D over denominator 2 end fraction close parentheses
          = sin C - sin D = L.H.S.
Hence,  space space space space sin space straight C minus space sin space straight D space equals space 2 cos fraction numerator straight C plus straight D over denominator 2 end fraction sin fraction numerator straight C minus straight D over denominator 2 end fraction

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