"let " sin(pi/3+x)-cos(pi/6+x) " be equal to 'k' "let sin(π3+x)−cos(π6+x) be equal to 'k'
sin(a+b)=sin a*cos b +cos a * sin bsin(a+b)=sina⋅cosb+cosa⋅sinb
sin(pi/3+x)=sin( pi/3)*cos x+cos(pi/3)*sin xsin(π3+x)=sin(π3)⋅cosx+cos(π3)⋅sinx
sin(pi/3+x)=sqrt 3/2*cos x+1/2*sin xsin(π3+x)=√32⋅cosx+12⋅sinx
cos(a+b)=cos a*cos b-sin a*sin bcos(a+b)=cosa⋅cosb−sina⋅sinb
cos(pi/6+x)=cos(pi/6)*cos x-sin(pi/6)*sin xcos(π6+x)=cos(π6)⋅cosx−sin(π6)⋅sinx
cos(pi/6+x)=sqrt 3/2*cos x-1/2* sin xcos(π6+x)=√32⋅cosx−12⋅sinx
sqrt 3/2*cos x+1/2*sin x-(sqrt 3/2*cos x-1/2* sin x)√32⋅cosx+12⋅sinx−(√32⋅cosx−12⋅sinx)
cancel(sqrt 3/2*cos x)+1/2*sin x-cancel(sqrt 3/2*cos x)+1/2 *sin x=k
1/2*sin x+1/2*sin x=k
1/2 sin x(1+1)=k
1/cancel(2) *cancel(2) sin x=k
sin x=k
sin(pi/3+x)-cos(pi/6+x)=k
sin(pi/3+x)-cos(pi/6+x)=sin x
