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junior college 1 | H2 Maths
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Complex number
Date Posted: 6 years ago
Views: 314
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Date Posted:
6 years ago
My "j" is the "i"...
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Correct?
Date Posted:
6 years ago
Why suddenly have pie k after removing sin
For positive angles, sin0=0, sin(pi)=0, sin(2pi)=0. Second round, sin(3pi)=0, sin(4pi)=0. No limit is given, so the number of rounds go on. Same for the negative angles...
So, in general, sin(k*pi)=0 where k can take 0, +/-1, +/-2, +/-3, ...
Question asked for positive, so take only the k that make n positive.
So, in general, sin(k*pi)=0 where k can take 0, +/-1, +/-2, +/-3, ...
Question asked for positive, so take only the k that make n positive.
In secondary level, we will present the working as follows:
Basic angle =0 or pi
Hence (pi /6)(n+1)= pi, 2pi, 3pi.
n = ...
But in JC, you should use k and state k=0, +/-1...
Basic angle =0 or pi
Hence (pi /6)(n+1)= pi, 2pi, 3pi.
n = ...
But in JC, you should use k and state k=0, +/-1...
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