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Question Number 125833 by bramlexs22 last updated on 14/Dec/20
4sin(2π7)+sec(π14)cot(π7)?
Answered by Dwaipayan Shikari last updated on 14/Dec/20
4sin2π7cosπ14sinπ7+sinπ7cosπ14cosπ7=2(cosπ7−cos3π7)cosπ14+sinπ7cosπ14cosπ7=cosπ14−cos3π14−cos5π14+cosπ2+cos5π14cosπ14cosπ7sinπ7=cos5π14=cosπ14−cos3π7cosπ14cosπ7=2(cosπ14cosπ7cosπ14cosπ7)=2
Answered by liberty last updated on 14/Dec/20
letπ14=xthen4sin4x+secxcot2x=4sin4xcosx+1cosx(cos2xsin2x)=4sin4xcosxsin2x+sin2xcosxcos2x=−2(cos6x−cos2x)cosx+sin2xcosxcos2x=2cos2xcosx−2cos6xcosx+sin2xcosxcos2x=2cos2xcosx−cos7x−cos5x+sin2xcosxcos2x[weknowthatcos7x=cosπ2=0∧cos5x=cos5π14=sin2π14]thuswegetcos7x−cos5x+sin2x=0so2cos2xcosx+0cosxcos2x=2.
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