End points of a graph?

Question

Can someone please help me out? I get stuck at the end...

y= 3sin[2(x+(pi/3))] , domain [-pi, pi]

I get to
x=pi y=3sin(4pi/3)
but I don't quite know how to simplify it futher.
Can you please show working out, Thanks!

Answer 1

4pi/3 is pi/3 more than pi, which is a straight angle, so it has the same proportions as an angle of pi/3, except that its sine is negative. Now, pi/3 is the same as an angle of 60 degrees, which has an adjacent side of 1/2 and an opposite side of sqrt(3)/2 when the length of the hypotenuse is 1, so sin(4pi/3) = -sqrt(3)/2. If you want to multiply it by three, you end up with -(3/2)sqrt(3).

Answer 2

just substitute the value -pi in the given equation.
ie, y=3 sin[2(-pi+pi/3)] = 3 sin(- 4pi/3)
Then you get y = --2.598 ; when rounded off to 3 digits
This is the initial point of the graph w.r.t the argument -pi
Now, put pi in the equation, you get y= 3sin[2(pi+pi/3)]
=3 sin(8pi/3) = 2.598
This is the final point of the graph w.r.t. the argument pi

Answer 3

I am not sure how you get to your answer but the brute force method for finding sin(4Pi/3) without having to worry about its sign is to use the identity sin(A+B)=sinAcosB+cosAsinB

sin(4Pi/3)= si(Pi+Pi/3)

=sinPicosPi/3+cosPicosPi/3

=-1*(Sqrt3)/2