Complex De Moivre's & Trig Question (1 Viewer)

[pixiefeathers]

Hi, I attempted part b of this question (Cambridge X2 Exercise 3B Q14) and thought I got the answers but they were incorrect. I think I got it wrong when I equated 8θ = 0, π, 2π, etc. but I'm not sure. Would someone be able to tell me how I went wrong? I have attached my solutions and the actual solutions. I'd really appreciate some help, thanks.

Question:


My solution (part a is correct):



Cambridge solution:

 

[tywebb]

 

[tywebb]

As you can see there are 6 distinct roots and none repeated.

0 is not one of those roots.

However you express the trigonometric solutions, they have to be distinct and none repeated.

2 of your solutions, 2sin(3π/8) and 2sin(5π/8) are the same

2sin(π/2) = 2 is also not one of the solutions.

Another way to express them is ±2sin(π/8), ±2sin(π/4), ±2sin(3π/8)

 

[pixiefeathers]

Thank you! Is it that when I solve sin8θ = 0, I should always put 8θ = ±π, ±2π, ±3π, etc., instead of π, 2π, 3π, 4π, etc.? Because I seem to only be able to get the correct answers if I put the former not the latter.

 

[tywebb]

Well it's more about having the correct number of solutions and ones that satisfy the original equation. With trigonometric solutions there are many ways to express the solutions. Sometimes the domain will also play a part.

In this case you could use 8θ=nπ for integers n. But which integers you chose should give 6 distinct solutions satisfying the original equation.

The most obvious one is n=±1, ±2, ±3

But another way could be that n=1, 2, 3, 9, 10, 11

So the solutions could also be expressed as 2sin(π/8), 2sin(π/4), 2sin(3π/8), 2sin(9π/8), 2sin(5π/4), 2sin(11π/8)

This is because 2sin(9π/8) = -2sin(π/8), 2sin(5π/4) = -2sin(π/4), 2sin(11π/8) = -2sin(3π/8)

 

[pixiefeathers]

Right, I will remember to test those first. Thank you so much for your help!