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

Question

4475 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-26T19:23:02

$=sqrt3/2 cosx-1/2 sinx-sqrt3/2sinx-1/2 cosx$

Use Properties:
$cos(A+B)=cosAcosB-sinAsinB$
$A=x,B=pi/6$

$sin(A+B)=sinAcosB+cosAsinB$
$A=x,B=pi/6$

$cos(x+pi/6)-sin(x+pi/6)=cosxcos(pi/6)-sinxsin(pi/6)-[sinxcos(pi/6)+cosxsin(pi/6)]$

$=cosxcos(pi/6)-sinxsin(pi/6)-sinxcos(pi/6)-cosxsin(pi/6)]$

$=sqrt3/2 cosx-1/2 sinx-sqrt3/2sinx-1/2 cosx$

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