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

Question

Asher Friedman · Accepted Answer

#=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#

Quinn Anderson · Answer

undefined Cos(x+pi/6) - sin(x+pi/6) can be simplified using trigonometric identities, specifically the angle addition formulas for cosine and sine:

Cos(x+pi/6) = cos(x)cos(pi/6) - sin(x)sin(pi/6)
Sin(x+pi/6) = sin(x)cos(pi/6) + cos(x)sin(pi/6)

Substituting these expressions into Cos(x+pi/6) - sin(x+pi/6) and simplifying gives:

cos(x)cos(pi/6) - sin(x)sin(pi/6) - (sin(x)cos(pi/6) + cos(x)sin(pi/6))
= cos(x)cos(pi/6) - sin(x)sin(pi/6) - sin(x)cos(pi/6) - cos(x)sin(pi/6)
= (cos(x)cos(pi/6) - sin(x)cos(pi/6)) - (sin(x)sin(pi/6) + cos(x)sin(pi/6))
= cos(pi/6)(cos(x) - sin(x)) - sin(pi/6)(sin(x) + cos(x))

Then, applying the values of cos(pi/6) and sin(pi/6) (which are sqrt(3)/2 and 1/2 respectively):

= (sqrt(3)/2)(cos(x) - sin(x)) - (1/2)(sin(x) + cos(x))
= (sqrt(3)/2)cos(x) - (sqrt(3)/2)sin(x) - (1/2)sin(x) - (1/2)cos(x)
= (sqrt(3)/2)cos(x) - (1/2)cos(x) - (sqrt(3)/2)sin(x) - (1/2)sin(x)

Finally, combining like terms gives the simplified expression:

= (sqrt(3)/2 - 1/2)cos(x) - (sqrt(3)/2 + 1/2)sin(x)
= (sqrt(3) - 1)/2 cos(x) - (sqrt(3) + 1)/2 sin(x)