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)??

Question

anonymous · Answer

what?

anonymous · Answer

i expect you used the compound angle formula to do this problem
and got
Asin (x + c)  = 
Asin x cos c + A cos x sinc
and compare this identity to
                          A sin x  + A cos x * √3

so A = √(1 +3) = 2
and tan c = √3
so c = π/3

giving 2 sin (x + π/3)

anonymous · Answer

I got your point,but if I use the method like following I got the different answer: As sinx+√3cosx=asinkx+bcoskx,then a=1,b=√3,k=1,A=√(a^2+b^2)=√(1+3)=2,tankO(fai)=-b/a=-√3,then sinx+√3cosx=2sin(x-2/3π)

dumbcow · Answer

i agree with jimmyrep.
however
tan(pi/3) = tan(-2pi/3)
therefore your answer is technically correct as well

dumbcow · Answer

wait nevermind, A would have to be -2 though

dumbcow · Answer

im not following your method...

anonymous · Answer

this is not a different answer this is the same answer