Question

From Euler’s relation eiθ = cosθ + isinθ,

Answer 1

Answer:

(a) The graphic representation is in the attached figure.

(b) $\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$.

(c) $\sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}$.

Step-by-step explanation:

(a) Given a complex number $e^{i\theta}$ we know, from Euler's formula that $e^{i\theta} = \cos(\theta)+i\sin(\theta)$. So, it is not difficult to notice that

$|e^{i\theta}|^2 = \cos^2(\theta)+\sin^2(\theta) =1$

so it is on the unit circumference. Also, notice that the Cartesian representation of the complex number is $(\cos(\theta), \sin(\theta))$.

Now,

$e^{-i\theta} = \cos(\theta)+i\sin(-\theta) = \cos(\theta)-i\sin(\theta)$.

Notice that $e^{-i\theta}$ has the same modulus that $e^{i\theta}$, so it is on the unit circumference. Beside, its Cartesian representation is $(\cos(\theta), -\sin(\theta))$.

So, the points $(\cos(\theta), \sin(\theta))$ and $(\cos(\theta), -\sin(\theta))$ are symmetric with respect to the X-axis. All this can be checked in the attached figure.

(b) Notice that

$e^{i\theta} + e^{-i\theta} = \cos(\theta)+i\sin(\theta) + \cos(\theta)-i\sin(\theta) = 2\cos(\theta)$

Then,

$\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$.

(c) Notice that

$e^{i\theta} - e^{-i\theta} = \cos(\theta)+i\sin(\theta) - \cos(\theta)+i\sin(\theta) = 2i\sin(\theta)$

Then,

$\sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}$.