A Complex Cofunction Equivalency For A Complex Exponential

Euler's Formula states that for any real number $x$:

\begin{equation}\label{eulers_formula} e^{ix} = \cos(x) + i\sin(x) \end{equation}

This can be shown using the Taylor series expansions for the exponential, sine, and cosine functions, and using the complex definition of $i$.

Logarithm of a Negative Number

$$ \log_2(-16) = \log_2(16) + \log_2(-1) = 4 + \frac{\ln(-1)}{ln(2)}= 4 + \frac{i\pi}{ln(2)} $$

The last step above is solved by using Euler's fomula for the imaginary exponent, $e^{ix} = \cos(x) + i\sin(x)$, which is derived using the Taylor series, in a subsequent article.

$$ e^{ix} = \cos(x) + i\sin(x) = -1 \rightarrow x=\pi + 2k\pi, k \in \mathbb{Z} \rightarrow \ln(-1)=i\pi $$

$$ y=2^x \rightarrow \ln(y) = x\ln(2) \rightarrow e^{\ln(y)} = e^{x\ln(2)} \rightarrow 2^x = e^{x\ln(2)} $$
$$ 2^{4+\frac{i\pi}{ln(2)}} = 16\left(2^{\frac{i\pi}{ln(2)}}\right) = 16e^{i\pi} = -16 $$

The Angle Addition Formula

The Angle Addition Formula states that for any angles $\theta$ and $\phi$:

\begin{equation}\label{angle_addition_sine} \sin(\theta + \phi) = \sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi) \end{equation}

Using Euler's formula, we can also express the sine and cosine functions in terms of complex exponentials:

$$ \sin(x) = \frac{e^{ix} - e^{-ix}}{2i}, \quad \cos(x) = \frac{e^{ix} + e^{-ix}}{2} $$

Now, let's derive the angle addition formula for sine using these expressions:

$$ \sin(\theta + \phi) = \frac{e^{i(\theta + \phi)} - e^{-i(\theta + \phi)}}{2i} $$ We can expand the exponentials using the properties of exponents: $$ = \frac{e^{i\theta}e^{i\phi} - e^{-i\theta}e^{-i\phi}}{2i} $$ Now, we can substitute the expressions for sine and cosine: $$ = \frac{(\cos{\theta} + i\sin{\theta})(\cos{\phi} + i\sin{\phi}) - (\cos{\theta} - i\sin{\theta})(\cos{\phi} - i\sin{\phi})}{2i} $$ Factor out the terms: $$ = \frac{(\cos{\theta}\cos{\phi} + i\sin{\theta}\cos{\phi} + i\sin{\phi}\cos{\theta} - \sin{\theta}\sin{\phi}) - (\cos{\theta}\cos{\phi} - i\sin{\theta}\cos{\phi} - i\sin{\phi}\cos{\theta} + \sin{\theta}\sin{\phi})}{2i} $$ Simplify the expression: $$ = \frac{2i(\sin{\theta}\cos{\phi} + \cos{\theta}\sin{\phi})}{2i} $$ Finally, we arrive at the angle addition formula for sine: $$ = \sin{\theta}\cos{\phi} + \cos{\theta}\sin{\phi} $$

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