How do you verify #Cos(x+(pi/6))- cos(x-(pi/6))=1#?

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Answer 1 · 2016-09-24T12:40:25

#cos(x+pi/6)-cos(x-pi/6)=-sinx# and not #1#.

It is #1#, only when #-sinx=1# i.e. #x=(3pi)/2#

As #cos(A+B)=cosAcosB-sinAsinB# and

#cos(A-B)=cosAcosB+sinAsinB#

Hence #cos(A+B)-cos(A-B)=-2sinAsinB#

Hence #cos(x+pi/6)-cos(x-pi/6)=-2sinxsin(pi/6)#

= #-2sinx xx1/2=-sinx#

Hence #cos(x+pi/6)-cos(x-pi/6)=-sinx# and not #1#.

It is #1#, only when #-sinx=1# i.e. #x=(3pi)/2#