妙妙貓

I attached my solutions for (i), (ii) and (A) and (B) of (iii), but not sure how to approach (C). Can someone please give some hints?

妙妙貓

Oh yes, thank you. I used cosine formula to find the modulus and solved it.

妙妙貓

Any suggestions for these two?

For Q5, I did the part (a), (b) and (c), but unsure how to prove the part (d).

For Q17, I think a possible way to do the question is to substitute all the values into the original equation and show it equals 0, but there should probably be a faster method.

妙妙貓

For Question 5, this is where I got so far. I know there is formula for product of 2 cosines cosAcosB=(1/2)[cos(A-B)+cos(A+B)]. And for sum of 2 cosines, I know cosA+cosB=2cos[(A+B)/2]cos[(A-B)/2], but not sure how it is applicable to the question.

For Question 17, I have learnt roots of unity, but still not sure how to approach the question.

妙妙貓

Sorry, I think the image didn't send

[ 本帖最後由 妙妙貓 於 2022-3-10 04:27 PM 編輯 ]

妙妙貓

Oh yes, I got Q5. Originally, I didn't expand the cis into cos+isin and didn't notice that things would cancel.

And thank you for the hint for Q17, I solved it too.

妙妙貓

Could you please also give a hint about this one?

It defines w as a non-real cube root of unity, and I notice that z^2+z+1 is part of the factorisation of z^3-1=0 and if w is a non-real cube root of unity, then (substituting z as w), it will equal 0. But other than that, I have no idea about this question.

妙妙貓

I proved that P_n(w)=0. But I'm a little confused as to why we can substitute the w back to x. In my working out, there is:

Because P_n(w)=0, (x-w) is a factor of P_n(x).

Because w^2+2+1=0 (x-w)=(x+1+w^2)

And then I can just substitute w as x to get "therefore (x^2+x+1) is a factor of P_n(x)"? Or did I misunderstand your meaning?

妙妙貓

Oh thanks. I understand now. I used w^2 as the second factor and it worked.