MandysNotes

We can find the half angle formula for cosine by starting with the double angle formula:

\[ \cos{(2\alpha)} = \cos^{2}{\alpha} - \sin^{2}{\alpha} \ . \]

Making the substitution:

\[ \gamma = 2\alpha ,\]

leads to the equation:

\[ \cos{\gamma} =  \cos^{2}{(\frac{\gamma}{2})} - \sin^{2}{(\frac{\gamma}{2})} . \]

From the definition of cosine:

\[ \cos{\theta} =\frac{e^{i\theta} + e^{-i\theta}}{2} \]

we find:

\[ 2\cos{(\alpha + \beta)} = e^{i(\alpha + \beta)} + e^{-i(\alpha + \beta)} .\]

Consider any complex number \[ z, \ \]with modulus

 

\[ \sqrt{z\bar{z} } = |z| = r \ ,\]

where \[ r \in \mathbb{R^{+}}\ . \]

Note that \[ \frac{z}{r} \ \]would be a complex number with modulus one:

\[ | \frac{z}{r} | = | \frac{z}{|z|} | = 1. \]

Euler's formula makes life much easier when it comes to deriving trigonometric identities.

 

Note, first of all, that \[ |e^{i\alpha}| = 1, \]gives us:

\[ \left( \text{Re} (e^{i\alpha}) + i\text{Im}(e^{i\alpha}) \right) \left( \text{Re}(e^{i\alpha}) – i\text{Im}(e^{i\alpha}) \right) \]

\[ = \left( \cos{\alpha} + i\sin{\alpha} \right) \left( \cos{\alpha} - i\sin{\alpha} \right) \]

\[ = \cos^{2}{\alpha} + \sin^{2}{\alpha} = 1. \]

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