Question

Using de Moivre's identity,
find the values of a and b in the equation, such that the equation is valid.
\[\cos^6 \theta + \sin^6 \theta + a(\cos^4 \theta + \sin^4 \theta) + b = 0\]

Check values for a and b.
Also says that it helps to write cos(theta) and sin(theta) in terms of e^i(theta).

No idea how to start this off... anyone that can help?

Answer 1

This isn't using de Moivre's theorem, but my first stab at this is to write everything in terms of \( \cos^2 \theta \). Write \( c = \cos \theta \), then your equation is
\[ c^6 + (1-c^2)^3 + a(c^4 + (1-c^2)^2) + b = 0 \]
i.e.
\[ 2c^6 + (2a-3)c^4 + (3- 2a) c^2 + (a + b + 1) = 0 \]
Now you can use Descartes rule of signs to characterize what kind of a and b will enable this equation to be satisfied; we have a third order polynomial in c^2 which must always be positive.

I'll be curious to see how to tackle this with deMoivre.

Answer 2

Yeah I'll take a look at it and see what I can get done to it.
It says specifically to use de Moivre's identity though and to convert \[\cos (\theta) and \sin (\theta)\]into terms of \[(e^{i \theta}) and (e^{-i \theta})\]

From what I know:
\[\cos \theta = {e^{i \theta} + e^{-i \theta} \over 2}\]
\[\sin \theta = {1 \over i } . {e^{-i \theta} - e^{i \theta} \over 2}\]

Answer 3

I think I have an idea about how to tackle this using de Moivre's formula. If we rewrite \(\cos^6(x)=\frac{e^{ix}+e^{-x}}{2}\), then we can expand this as:
\[\cos^6{x}=\frac{1}{2^6}(e^{6ix}+6e^{4x}+15e^{2ix}+20+15e^{-2ix}+6e^{-4ix}+e^{-6ix})\]
Here you can apply de Moivre's identity and find that \[\cos^6{x}=\frac{1}{2^6}(2\cos(6x)+12\cos(4x)+30\cos(2x)+20).\]

Answer 4

Using the similar approach:
\[\sin^6(x)=\frac{1}{2^6}(-2\cos(6x)+12\cos(4x)-30\cos(2x)+20).\]

Answer 5

oh, I see.

Answer 6

I think the idea is clear now, you will get a linear equation in cos which will be easier to solve.

Answer 7

In fact this method is often used to integrate higher powers of cos or sin, for instance \(cos^6(x)\).

Answer 8

Yeah I was on my way to coming up with the same thing,
Nice one mate!
Ha that's probably what I'm moving onto next (y).