Complex number and trig -- past paper question

[cccc]

Hi, I have a slight query on the following question. This is from a higher level paper (i have attached both the question and the answer) 

 

For part (c) I don't know how they could exchange 8pi/5 with 2 pi/5 and don't quite understand the sign changes. This goes with 6pi/5 to be replaced with 4pi/5

Because if you draw out the CAST diagram cos 8 pi/5 should be in the A qudrant (from 0 to pi/2) which means that both cosine and sine should be positive? and not cosine being positive and sine being negative as the answer suggested. 

Any advice would be great! Thanks. 

 

[kw0573]

While I was revising for HL Math this was one of the most difficult questions I encountered. Just to clarify, Q1(A) is 0 to pi / 2 (or 2.5 pi / 5 ), Q2 (S) is pi/2 to pi (or 5 pi/5). Q3 (T) is pi to 3pi/2 (or 7.5 pi / 5). and Q4(C) is 3pi/2 to 2pi (or 10 pi / 5). So 8 pi /5 is in the C or fourth quadrant. 

For the change of argument question. cos is horizontal component of a complex number, and remain the same the the complex number is reflected across the real axis (to become the complex conjugate). For the complex number cos (6 pi / 5) + i sin (6 pi/5), angle 6 pi  / 5 reflected across the real axis is 4 pi / 5 (from pi + pi/5 to pi - pi/5). 
However sin is vertical component so when it reflects across the real axis the sign is reversed. So we would have -(sin 4pi / 5) as opposed to just sin (4pi/5). 
Similar argument for 8 pi / 5. From pi + 3pi/5 it reflects to become pi - 3pi/5. The sin component would equally be switching signs.

 

[cccc]

12 hours ago, kw0573 said:

However sin is vertical component so when it reflects across the real axis the sign is reversed. So we would have -(sin 4pi / 5) as opposed to just sin (4pi/5). 
Similar argument for 8 pi / 5. From pi + 3pi/5 it reflects to become pi - 3pi/5. The sin component would equally be switching signs.

 

 

@kw0573Thank you very much! However, my query is also this: I understand what u said abt the reflection on the real axis and the signs, but the 8 pi/5 thing corresponds to 3 pi/5 and not 2pi/5 so cannot cancel out with the 2pi/5 one in the first term? 

Edited April 8, 2016 by cccc

[kw0573]

@cccc 8pi/5 is pi + 3pi/5. After reflection it is pi - 3 pi/5 = 2pi / 5. 
You can also think of in terms of 0 or 2pi. 8 pi/5 is in the same position as 2pi  - 2pi/5 or just -2pi / 5. This is reflected to be +2pi/5.

[Jai Chandak]

Which paper was this question from?