0
$\begingroup$

De Moivre's formula: $(\cos(\phi) + \sin(\phi)i)^n = \cos(n\phi)+\sin(n\phi)i$

I want to prove, derived from the formula: $\cos(3\phi) = 4\cos(\phi)^3 -3\cos(\phi)$

In my attempt I used the formula for $(a+b)^3$ and De Moivres and

$\sin(\phi)^2 = \cos(\phi)^2 + 1$

It looks complicated at first glance but it isn't, still im stuck at the end and im not sure how to proceed. Either I made the same mistake(s) multiple times (tried this a few times) or I have to work with the last part, which I don't know how to.

$(\cos(\phi) + \sin(\phi)i)^3 = (\cos(\phi) + \sin(\phi)i)^3 \Leftrightarrow$

$\cos(3\phi) + \sin(3\phi)i = \cos(\phi)^3 + \cos(\phi)^2\sin(\phi)i + \cos(\phi)\sin(\phi)^2i^2 + \sin(\phi)^3i^3\Leftrightarrow$

$\cos(3\phi) = \cos(\phi)^3 + \cos(\phi)^2sin(\phi)i + \cos(\phi)\sin(\phi)^2i^2 + \sin(\phi)^3i^3 - \sin(3\phi)i\Leftrightarrow$

$\cos(3\phi) = \cos(\phi)^3 + \cos(\phi)^2\sin(\phi)i + \cos(\phi)(\cos(\phi)^2 + 1)i^2 + (\cos(\phi)^2 + 1)(\sin(\phi))i^3 - \sin(3\phi)i\Leftrightarrow$

$\cos(3\phi) = \cos(\phi)^3 + \cos(\phi)^2\sin(\phi)i - \cos(\phi)^3 - \cos(\phi) + (\cos(\phi)^2\sin(\phi) + \sin(\phi))i^3- \sin(3\phi)i\Leftrightarrow$

$\cos(3\phi) = \cos(\phi)^3 + \cos(\phi)^2\sin(\phi)i - \cos(\phi)^3 - \cos(\phi) - (\cos(\phi)^2\sin(\phi) + \sin(\phi))i- \sin(3\phi)i\Leftrightarrow$

$\cos(3\phi) = \cos(\phi)^3 + \cos(\phi)^2sin(\phi)i - \cos(\phi)^3 - \cos(\phi) - \cos(\phi)^2\sin(\phi) i- \sin(\phi)i- \sin(3\phi)i\Leftrightarrow$

$\cos(3\phi) = - \cos(\phi) i- \sin(\phi)i- \sin(3\phi)i$

  • 2
    You lost the binomial coefficients (the two $3$s, in this case) right where you expanded. – 2017-02-09
  • 0
    Oh yes, I missed the 3's. But $\sin(3\phi)i$ moved over to the right side – 2017-02-09

2 Answers 2

5

Sorry but please make adjustments to your $\cos$ and $\sin$ use a '\cos' and '\sin' (I just find yours unreadable)

Anyways this is what I would do

$$ (\cos(\phi) + \sin(\phi)i)^n = \cos(n\phi)+\sin(n\phi)i$$

Let $n=3$

$$ (\cos(\phi)+\sin(\phi)i)^3=\cos(3\phi)+\sin(3\phi)i$$

$$ \cos^3(\phi)+3\cos^2(\phi)\sin(\phi)i-3\cos(\phi)\sin^2(\phi)-\cos(\phi)\sin^3(\phi)i=\cos(3\phi)+\sin(3\phi)i $$

Compare real parts:

$$ \cos(3\phi)=\cos^3(\phi)-3\cos(\phi)\sin^2(\phi)$$

$$ \cos(3\phi)=\cos^3(\phi)-3\cos(\phi)(1-\cos^2(\phi))$$

$$ \cos(3\phi)=\cos^3(\phi) - 3\cos(\phi)+3\cos^3(\phi) $$

$$ \Longrightarrow \cos(3\phi)=4\cos^3(\phi)-3\cos(\phi)$$

1

The formula for $(a+bi)^3$ is \begin{align} (a+bi)^3 &=a^3+3a^2(bi)+3a(bi)^2+(bi)^3 \\ &=a^3+3a^2bi-3ab^2-b^3i \\ &=(a^3-3ab^2)+(3a^2b-b^3)i \end{align} Therefore the real part of $(\cos\phi+i\sin\phi)^3$, that is $\cos3\phi$, is, substituting $a=\cos\phi$ and $b=\sin\phi$, $$ \cos3\phi=\cos^3\phi-3\cos\phi\sin^2\phi= \cos^3\phi-3\cos\phi(1-\cos^2\phi) $$