I need to know a bit about replacing the trig functions in exponents with simpler algebraic equations which can then be reduced with logarithms. For example, an answer to this question said that
$$e^{\sin(x)} = e^{\frac{1}{2i}(e^{ix} – e^{-ix})}$$
applies, which would do the trick. As well as this, using a few basic log rules it could be adjusted to deal with any base as well. I do have a few questions though.
Firstly, how does this work? It looks like it has a fair bit to do with Euler's identity but I don't understand the exact derivation.
Next, how do I generalize it (For example, with the exponent as $4*sin^2(3x)$)? I honestly don't know if it barely involves changing numbers or if it doesn't apply at all anymore.
Finally, is this transferrable to the other trig functions? If it's connected to Euler's identity as I think it is, it should be easy to change it to apply to cosine instead, and then the two together may somehow be used to derive the transformation (if I can call it that) to tan.
Thank you so much for your help!
Best Answer
The derivation of the definitions of $\sin x$ and come from subtracting Euler's formulas. \begin{align} e^{ix}&=\cos x+i\sin x\\ e^{-ix}&=\cos x-i\sin x\\ \implies\sin x&=\frac1{2i}(e^{ix}-e^{-ix}) \end{align} It therefore follows that $$\exp(\sin x)=\exp\left(\frac1{2i}(e^{ix}-e^{-ix})\right).$$
For your example, we can write \begin{align} 4\sin^2(3x)=4(\sin(3x))^2&=4\left(\frac1{2i}(e^{3ix}-e^{-3ix})\right)^2\\ &=-(e^{3ix}-e^{-3ix})^2\\ &=2-e^{6ix}-e^{-6ix}\\ \implies\exp(4\sin^2(3x))&=\exp(2-e^{6ix}-e^{-6ix}).\\ \end{align}