METHODS VCAA 2016

[minhalgill]

Hi.
In 6(a), in solving sin(-2pi/3),  it becomes sin(-pi+pi/3), which becomes sin(pi/3) because we get rid of the -pi. why is this wrong? can you only get rid of pi? not -pi?

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Hi.
In 6(a), in solving sin(-2pi/3),  it becomes sin(-pi+pi/3), which becomes sin(pi/3) because we get rid of the -pi. why is this wrong? can you only get rid of pi? not -pi?

sin(-2pi/3) = sin(-pi/3) = -sqrt3/2 because sine is negative in the 3rd and 4th quadrant