How do you prove $Cos(x+pi/6)-sin(x+pi/6)$ ?

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Answer 1 · 2017-01-17T18:09:16

$=sqrt3 /2 cos x - 1/2 sinx - sqrt3 / 2 sinx - 1/2 cos x$

Use the formulas: $cos(A+B)=cosAcosB-sinAsinB$ and
$sin(A+B)=sinAcosB+cosAsinB$

$cos(x+pi/6)-sin(x+pi/6)=[cosxcos(pi/6)-sinx sin (pi/6)]-[sinx cos(pi/6)+cos x sin (pi/6)]$

$=[cosx (sqrt3 /2)-sinx (1/2)]-[sinx (sqrt3 /2)+cos x sin(pi/6)]$

$=sqrt3 /2 cos x - 1/2 sinx - sqrt3 / 2 sinx - 1/2 cos x$

How do you prove Cos(x+pi/6)-sin(x+pi/6)Cos(x+pi/6)-sin(x+pi/6)?