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

1 Answer
Bdub
Jan 17, 2017

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

Explanation:

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#

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