Sets

Sets

Question

Using principle of mathematical induction, prove that

cos space straight alpha space space cos space 2 straight alpha space space cos space 4 straight alpha space............... cos left parenthesis 2 to the power of straight n minus 1 end exponent straight alpha right parenthesis space equals space fraction numerator sin space 2 to the power of straight n straight alpha over denominator 2 to the power of straight n sinα end fraction for all straight n space element of space straight N

Answer

Let P(n) : space space cosα space cos 2 straight alpha space cos 4 straight alpha space.......... cos left parenthesis 2 to the power of straight n minus 1 end exponent straight alpha right parenthesis space equals space fraction numerator sin left parenthesis 2 to the power of straight n straight alpha right parenthesis over denominator 2 to the power of straight n sinα end fraction
I.   For n = 1,
    straight P left parenthesis 1 right parenthesis space colon space cosα space equals space fraction numerator sin space 2 straight alpha over denominator 2 sin space straight alpha end fraction rightwards double arrow space cosα space equals space fraction numerator 2 sinα space cosα over denominator 2 space sin space straight alpha end fraction rightwards double arrow space cosα space equals space cos space straight alpha

∴   P(1) is true
II.   Suppose the statement is true for n = m, straight m space element of space straight N
III.  P(m) : cos space straight alpha space space cos 2 straight alpha space cos 4 straight alpha space.......... space cos 2 to the power of straight m minus 1 end exponent space straight alpha space equals space fraction numerator sin space 2 to the power of straight m straight alpha over denominator 2 to the power of straight m sinα end fraction          ... (i)
       For n = m + 1,
       straight P left parenthesis straight m plus 1 right parenthesis space colon space cosα space space cos space 2 straight alpha space cos 2 squared straight alpha space............. space cos left parenthesis 2 to the power of straight m minus 1 end exponent straight alpha right parenthesis space cos space left parenthesis 2 to the power of straight m straight alpha right parenthesis space equals space fraction numerator sin left parenthesis 2 to the power of straight m plus 1 end exponent straight alpha right parenthesis over denominator 2 to the power of straight m plus 1 end exponent sin space straight alpha end fraction
      Now, cos space straight alpha space cos space 2 straight alpha space cos space 2 squared straight alpha space........... space cos left parenthesis 2 to the power of straight m minus 1 end exponent straight alpha right parenthesis space cos left parenthesis 2 to the power of straight m space straight alpha right parenthesis
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#6 {main}</pre>
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#6 {main}</pre>                                                  [By (i)]
             = fraction numerator 2 space sin left parenthesis 2 to the power of straight m straight alpha right parenthesis space cos left parenthesis 2 to the power of straight m straight alpha right parenthesis over denominator 2 to the power of straight m plus 1 end exponent sinα end fraction equals space fraction numerator sin left parenthesis 2.2 to the power of straight m straight alpha right parenthesis over denominator 2 to the power of straight m plus 1 end exponent sinα end fraction equals space fraction numerator sin left parenthesis 2 to the power of straight m plus 1 end exponent straight alpha right parenthesis over denominator 2 to the power of straight m plus 1 end exponent sinα end fraction

∴      P(m + 1) is true.
       P(m) is true rightwards double arrow P(m + 1) is true
Hence, by the principle of mathematical induction, P(n) is true for all straight n space element of space straight N.

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