Principle of Mathematical Induction

Principle of Mathematical Induction

Question

Show that:  sin squared straight pi over 8 plus sin squared fraction numerator 3 straight pi over denominator 8 end fraction plus sin squared fraction numerator 5 straight pi over denominator 8 end fraction plus sin squared fraction numerator 7 straight pi over denominator 8 end fraction space equals space 2

Answer

L.H.S. = space space sin squared straight pi over 8 plus sin squared fraction numerator 3 straight pi over denominator 8 end fraction plus sin squared fraction numerator 5 straight pi over denominator 8 end fraction plus sin squared fraction numerator 7 straight pi over denominator 8 end fraction
          equals space space sin squared straight pi over 8 plus sin squared fraction numerator 3 straight pi over denominator 8 end fraction plus sin squared open parentheses fraction numerator 4 straight pi over denominator 8 end fraction plus straight pi over 8 close parentheses plus sin squared open parentheses fraction numerator 4 straight pi over denominator 8 end fraction plus fraction numerator 3 straight pi over denominator 8 end fraction close parentheses
           = space space sin squared straight pi over 8 plus sin squared fraction numerator 3 straight pi over denominator 8 end fraction plus sin squared open parentheses straight pi over 2 plus straight pi over 8 close parentheses plus sin squared open parentheses straight pi over 2 plus fraction numerator 3 straight pi over denominator 8 end fraction close parentheses
           = sin squared straight pi over 8 plus sin squared fraction numerator 3 straight pi over denominator 8 end fraction plus cos squared straight pi over 8 plus cos squared fraction numerator 3 straight pi over denominator 8 end fraction       open square brackets because space space sin open parentheses straight pi over 2 plus straight x close parentheses equals cos space straight x close square brackets
          equals space open square brackets sin squared straight pi over 8 plus cos squared straight pi over 8 close square brackets plus open square brackets sin squared fraction numerator 3 straight pi over denominator 8 end fraction plus cos squared fraction numerator 3 straight pi over denominator 8 end fraction close square brackets
           = 1 + 1                                                        (∵  sin squared straight x plus cos squared straight x space equals space 1)
           = 2 = R.H.S.
Hence, sin squared straight pi over 8 plus sin squared fraction numerator 3 straight pi over denominator 8 end fraction plus sin squared fraction numerator 5 straight pi over denominator 8 end fraction plus sin squared fraction numerator 7 straight pi over denominator 8 end fraction space equals space 2

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