# Simplifying this equation (euler's formula?)

I was reading some stuff earlier today, and I wasn't sure how they changed the exponentials to trigs in this expression:

$$C_1x^{-1/4}\exp\left(\frac{2}{3}ix^{3/2}\right)+C_2x^{-1/4}\exp\left(-\frac{2}{3}ix^{3/2}\right)$$

$=$

$$A_1x^{-1/4}\sin\left(\frac{2}{3}x^{3/2}\right)+A_2x^{-1/4}\cos\left(\frac{2}{3}x^{3/2}\right)$$

Does it have something to do with euler's formula?

And can I change $$C_1x^{-1/2}\exp\left(\frac{i}{2x^2}\right)+C_2x^{-1/2}\exp\left(-\frac{i}{2x^2}\right)$$ to $$A_1x^{-1/2}\sin\left(\frac{1}{2x^2}\right)+A_2x^{-1/2}\cos\left(\frac{1}{2x^2}\right)?$$

Yes, it does have something to do with Euler's formula. Briefly, $$\exp(iz)=\cos(z)+i\sin(z)$$

$$C_1x^{-1/4}\exp(\frac{2}{3}ix^{3/2})+C_2x^{-1/4}\exp(-\frac{2}{3}ix^{3/2})= C_1x^{-1/4}(\cos(\frac{2}{3}x^{3/2})+i\sin(\frac{2}{3}x^{3/2}))+C_2x^{-1/4}(\cos(\frac{2}{3}x^{3/2})-i\sin(\frac{2}{3}x^{3/2}))$$
And then we can expand out and collect like terms. We'll get that $A_1=iC_1-iC_2$ and $A_2=C_1+C_2$. A similar manipulation should work on your second question.
You can even go back the other way! We can express cosine and sine in terms of exp, if we're clever about how we change the sign of $z$:
$$\cos(z)= \frac{\exp(iz)+\exp(-iz)}{2}$$ $$\sin(z)= \frac{\exp(iz)-\exp(-iz)}{2i}$$