How do you verify #sin(x + pi/6) - cos(x + pi/3) = sqrt3 sin x#?

Question

25137 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-08T10:44:09

It is given that the LHS is #sin(x + pi/6) - cos(x + pi/3)#

Applying, #color(magenta)(sin(A+B) = sinAcosB + cosAsinB# and #color(red)(cos(A+B) = cosAcosB - sinAsinB#

#color(white)(dd#

#=> color(magenta)(sinxcos (pi/6) +cosxsin(pi/6))- (color(red)(cosxcos(pi/3) -sinxsin(pi/3))) #

#color(white)(dd#

#=> sinx(sqrt3/2) +cosx(1/2)- cosx(1/2) +sinx(sqrt3/2) #

#color(white)(dd#

#=>cancel 2xxsinx(sqrt3/cancel2) #

#=>sqrt3 sin x#

Answer 2 · 2018-04-08T10:55:19

See below.

Identities:

#color(red)bb(sin(A+B)=sinAcosB+cosAsinB)#

#color(red)bb(cos(A+B)=cosAcosB-sinAsinB)#

#LHS:#

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

#cos(x+pi/3)=cos(x)cos(pi/3)-sin(x)sin(pi/3)#

#sin(x+pi/6)-cos(x+pi/3)#

#sin(x)cos(pi/6)+cos(x)sin(pi/6)-[cos(x)cos(pi/3)-sin(x)sin(pi/3)]#

#sin(x)cos(pi/6)+cos(x)sin(pi/6)-cos(x)cos(pi/3)+sin(x)sin(pi/3)#

#cos(pi/6)=sqrt(3)/2#
#cos(pi/3)=1/2#
#sin(pi/6)=1/2#
#sin(pi/3)=sqrt(3)/2#

#sin(x)sqrt(3)/2+cos(x)(1/2)-cos(x)(1/2)+sin(x)(sqrt3)/2#

#2sin(x)sqrt(3)/2#

#sqrt(3)sin(x)#

#LHS-=RHS#