Principle Of Mathematical Induction

Question
CBSEENMA11014296

Prove the following:
cos open parentheses fraction numerator 3 straight pi over denominator 2 end fraction minus straight x close parentheses space equals space minus sinx,  space space sin open parentheses fraction numerator 3 straight pi over denominator 2 end fraction minus straight x close parentheses equals negative cosx,   tan open parentheses fraction numerator 3 straight pi over denominator 2 end fraction minus straight x close parentheses space equals space cotx


Solution

We know that: cos (x - y) = cosx cosy + sinx siny                                     ...(i)
                      sin (x - y) = sinx cosy - cosx siny                                       ...(ii)
                     cos open parentheses fraction numerator 3 straight pi over denominator 2 end fraction close parentheses equals 0                                                                    ...(iii)
                     sin open parentheses fraction numerator 3 straight pi over denominator 2 end fraction close parentheses equals negative 1                                                                 ....(iv)
Replacing x by open parentheses fraction numerator 3 straight pi over denominator 2 end fraction close parentheses and y by x in (i), we get
                    cos open parentheses fraction numerator 3 straight pi over denominator 2 end fraction minus straight x close parentheses space equals space cos open parentheses fraction numerator 3 straight pi over denominator 2 end fraction close parentheses space cosx space plus space sin open parentheses fraction numerator 3 straight pi over denominator 2 end fraction close parentheses space sinx space equals space 0 left parenthesis cosx right parenthesis space plus space left parenthesis negative 1 right parenthesis space sinx
                                          = - sinx                                         [By using (iii) and (iv)]
Hence,      cos open parentheses fraction numerator 3 straight pi over denominator 2 end fraction minus straight x close parentheses space equals space minus sinx
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#6 {main}</pre>  and y by x in (ii), we get
                    sin open parentheses fraction numerator 3 straight pi over denominator 2 end fraction minus straight x close parentheses space equals space sin open parentheses fraction numerator 3 straight pi over denominator 2 end fraction close parentheses space cosx space minus space cos open parentheses fraction numerator 3 straight pi over denominator 2 end fraction close parentheses space sinx space equals space left parenthesis negative 1 right parenthesis space cosx space minus space left parenthesis 0 right parenthesis space sinx
                                         = - cosx                                         [By using (iii) and (iv)]
Hence,      space sin open parentheses fraction numerator 3 straight pi over denominator 2 end fraction minus straight x close parentheses space equals space minus cosx
                space space tan open parentheses fraction numerator 3 straight pi over denominator 2 end fraction minus straight x close parentheses space equals space fraction numerator sin open parentheses begin display style fraction numerator 3 straight pi over denominator 2 end fraction minus straight x end style close parentheses over denominator cos open parentheses begin display style fraction numerator 3 straight pi over denominator 2 end fraction minus straight x end style close parentheses end fraction space equals space fraction numerator negative cosx over denominator negative sinx end fraction equals cotx
Hence,     tan open parentheses fraction numerator 3 straight pi over denominator 2 end fraction minus straight x close parentheses space equals space cotx
                    

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