How would you prove $\sin^2x + \cos^2x = 1$ using Euler's formula?
$$e^{ix} = \cos(x) + i\sin(x)$$
This is what I have so far:
$$\sin(x) = \frac{1}{2i}(e^{ix}-e^{-ix})$$
$$\cos(x) = \frac{1}{2} (e^{ix}+e^{-ix})$$
Answer
Multiply $\mathrm e^{\mathrm ix}=\cos(x)+\mathrm i\sin(x)$ by the conjugate identity $\overline{\mathrm e^{\mathrm ix}}=\cos(x)-\mathrm i\sin(x)$ and use that $\overline{\mathrm e^{\mathrm ix}}=\mathrm e^{-\mathrm ix}$ hence $\mathrm e^{\mathrm ix}\cdot\overline{\mathrm e^{\mathrm ix}}=\mathrm e^{\mathrm ix-\mathrm ix}=1$.
No comments:
Post a Comment