De moivres theorem stuff (1 Viewer)

constexpr

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Need a bit of help with 1 c... (ignore the blue underline)

I've got a means of solving i, ii, and iii partially but i'm not sure if it's the best approach.

Considering 1a and 1c(i) look fairly similar...
Where it asks to to solve for
I made
This gave me when you take the arccos of both sides...
Which gives This was one of the 4 correct roots, i got the other roots by simply adding and subtracting because the zero's repeat every 45 degrees

1cii i get a little stuck on finding the other roots. I can find one by rearranging 1ai)
I made
into


and finally it becomes


So now that i matched the equations... to solve 1ciii

I solved for


(take the arccos of both sides)
This is one the solutions but i don't know how to get the other roots, except for

Here are the answers...
1667632672314.png
Haven't started for iii, iv, not sure how to approach those.
Am I going about this all wrong? Any tips would be appreciated. cheers
 

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Your working for c) i) and ii) is correct, however, when finding values of 4θ, it is a better practice to write the solutions as 4θ = ±π/2, ±3π/2, ... (using i) as an example) as the solutions will occur in more quadrants than one. Once plugging in values for θ, you should also recognise that the expressions from c) i), ii) and iii) are also even functions, meaning that the ± sign should appear in front of the cos. Additionally, since the expressions are of degree 4, there can only be a maximum of 4 roots.

For c) iii), use the same method you used for ii) to obtain cos4θ = -1/2 and θ = 2π/3, 4π/3 then once again solving in each quadrant for θ and plugging back into a value for x, which will have exact values you can evaluate this time.

For c) iv), assuming you have already derived the expression for tan4θ in terms of tanθ, moving the 4x^3 - 4x term to the other side will yield 1-6x^2+x^4 = 4x-4x^3. Dividing both sides by 1-6x^2+x^4 to get 1 = (4x-4x^3)/(1-6x^2+x^4) and noticing the similarity between this expression and the expression for tan4θ, you can apply the substitution x = tanθ. You should end up with the result 1 = tan4θ.

Solving for θ; 4θ = π/4, 5π/4, -3π/4, -7π/4 ⇒ θ = π/16, 5π/16, -3π/16, -7π/16, and finally using the identity tan(-x) = -tanx, allows for tan(-3π/16) and tan(-7π/16) to be written as -tan(3π/16) and -tan(7π/16), respectively.
 

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