Tuesday, 22 August 2017

trigonometry - If A+B+C=π, prove that




If A+B+C=π, prove that
cos2A+cos2Bcos2C=2cosAcosBcosC.




ATTEMPT:
Given A+B+C=π,


A+B=πC

Taking "cos" on both sides

cos(A+B)=cosC.



Now,
L.H.S=cos2A+cos2Bcos2C=1+cos2A2+1+cos2B2cos2C=2+cos2A+cos2B2cos2C.



How should I complete now?



Answer



The identity




1cos2Acos2Bcos2C2cosAcosBcosC=0




can be seen as a re-writing of the relation cosC=cos(A+B) without sines of A and B:



cosC=cos(A+B)=sinAsinBcosAcosB



cosC+cosAcosB=sinAsinB

(cosC+cosAcosB)2=sin2Asin2B=(1cos2A)(1cos2B)

cos2C+2cosAcosBcosC+cos2Acos2B=1cos2Acos2B+cos2Acos2B

and the result follows.


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