Question

Hello,I have done the 1A Exercise and I got a problem in the 1a-6-a)Expressing(sinx+√3cosx) in the form Asin(x+c),the solution given was 2sin(x+π/3),but I thought it could be 2sin(x-2π/3),Is it could be 2sin(x-2π/3)??

Answer 1

(sinx+√3cosx)=(1+3)^1/2[sinx/(1+3)^1/2+{(3)^1/2.cosx}/(1+3)^1/2]=2[1/2.sinx+{(3)^1/2.cosx}/2}=2[cos(π/3)sinx+sin(π/3)cosx]=2sin(x+π/3)

Answer 2

sin x+√3cos x
(2/2)⋅(sin sx+√3cos x)
2((1/2)sin x+((√3)/2)cos x)
we can take many solutions of the system:
for example: We can take any angle a : sin a=(1/2); and cos a=((√3)/2), the angle is (π/6) or ((11π)/6) or -(π/6)

then 2⋅(sin asin x+cos acos x)=2⋅cos(x-a)
or we can take any angle a : cos a=(1/2); and sin a=((√3)/2), the angle is (π/3) or ((5π)/3) or -(π/3)
then 2⋅(cos asin x+sin acos x)=2⋅sin(x+a)