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

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

(I assume this is the form of Euler's formula you know. If not, Wikipedia provides a decent explanation. Please ask about this if it's unclear.)

So in your case, the manipulation is hidden a bit:

$$C_1x^{-1/4}\exp(\frac{2}{3}ix^{3/2})+C_2x^{-1/4}\exp(-\frac{2}{3}ix^{3/2})= C_1x^{-1/4}(\cos(\frac{2}{3}x^{3/2})+i\sin(\frac{2}{3}x^{3/2}))+C_2x^{-1/4}(\cos(\frac{2}{3}x^{3/2})-i\sin(\frac{2}{3}x^{3/2}))$$

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 $z$:

$$\cos(z)= \frac{\exp(iz)+\exp(-iz)}{2}$$ $$\sin(z)= \frac{\exp(iz)-\exp(-iz)}{2i}$$

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  • $\begingroup$ Thanks KReiser, I guess they were leaving out alot of the manipulation. $\endgroup$ – Gary Feb 24 '12 at 20:25

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