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Question

If sinπ14sin3π14sin5π14sin7π14sin9π14sin11π14sin13π14=1k, then k is equal to

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Solution

sinπ14sin3π14sin5π14sin7π14sin9π14sin11π14sin13π14
=sinπ14sin3π14sin5π14sinπ2sin(π5π14)sin(π3π14)sin(ππ14)
=sin2π14sin23π4sin25π14
=(sinπ14sin3π14sin5π14)2
=[cos(π2π14)cos(π23π14)cos(π25π14)]2
=[cos3π7cos2π7cosπ7]2=[cos(π4π7)cos2π7cosπ7]2=[cos2π7cosπ7cos4π7]2=[cosπ7cos2π7cos4π7]2=⎢ ⎢ ⎢sin8π723sinπ7⎥ ⎥ ⎥2=164⎢ ⎢ ⎢ ⎢sin(π+π7)sinπ7⎥ ⎥ ⎥ ⎥2=164k=64

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