Principle of Mathematical Induction

Principle of Mathematical Induction

Question

Prove that: 

2 sin squared open parentheses fraction numerator 11 straight pi over denominator 6 end fraction close parentheses plus cosec squared open parentheses fraction numerator 7 straight pi over denominator 6 end fraction close parentheses. cos squared open parentheses straight pi over 3 close parentheses space equals space 3 over 2

Answer

L.H.S. = 2 sin squared open parentheses fraction numerator 11 straight pi over denominator 6 end fraction close parentheses plus cosec squared open parentheses fraction numerator 7 straight pi over denominator 6 end fraction close parentheses. cos squared open parentheses straight pi over 3 close parentheses
        sin open parentheses fraction numerator 11 straight pi over denominator 6 end fraction close parentheses space equals space sin open parentheses fraction numerator 11 cross times 180 over denominator 6 end fraction close parentheses to the power of ring operator space equals space sin left parenthesis 330 degree right parenthesis space equals space sin left parenthesis 360 degree minus 30 degree right parenthesis space equals space sin open parentheses 2 straight pi minus straight pi over 6 close parentheses
Now,space space space space space 330 degree is in quadrant IV where sin space 330 degree is -ve.
rightwards double arrow          sin open parentheses fraction numerator 11 straight pi over denominator 6 end fraction close parentheses space equals space minus sin open parentheses straight pi over 6 close parentheses space equals space minus 1 half
Now,       sin open parentheses fraction numerator 7 straight pi over denominator 6 end fraction close parentheses space equals space sin open parentheses 7 over 6 cross times 180 close parentheses to the power of ring operator space equals space sin space 210 degree space equals space sin left parenthesis 180 degree plus 30 degree right parenthesis space equals space sin open parentheses straight pi plus straight pi over 6 close parentheses
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#6 {main}</pre> is in quadrant III where sin space 210 degree is -ve.
∴                 space space sin fraction numerator 7 straight pi over denominator 6 end fraction space equals space minus sin straight pi over 6 space equals space minus 1 half
rightwards double arrow                 space space space cosec fraction numerator 7 straight pi over denominator 6 end fraction space equals space minus 2
Also,                 cos open parentheses straight pi over 3 close parentheses space equals space 1 half
∴                L.H.S. = 2 open parentheses negative 1 half close parentheses squared plus left parenthesis negative 2 right parenthesis squared space open parentheses 1 half close parentheses squared space equals space 2 open parentheses 1 fourth close parentheses space plus space 4 open parentheses 1 fourth close parentheses space equals space 1 half plus 1 space equals space 3 over 2 space equals space straight R. straight H. straight S.

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