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. $x=(3pi)/2$

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