Answer:
(a) The graphic representation is in the attached figure.
(b) [tex]\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}[/tex].
(c) [tex]\sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}[/tex].
Step-by-step explanation:
(a) Given a complex number [tex]e^{i\theta}[/tex] we know, from Euler's formula that [tex]e^{i\theta} = \cos(\theta)+i\sin(\theta)[/tex]. So, it is not difficult to notice that
[tex]|e^{i\theta}|^2 = \cos^2(\theta)+\sin^2(\theta) =1[/tex]
so it is on the unit circumference. Also, notice that the Cartesian representation of the complex number is [tex] (\cos(\theta), \sin(\theta))[/tex].
Now,
[tex]e^{-i\theta} = \cos(\theta)+i\sin(-\theta) = \cos(\theta)-i\sin(\theta)[/tex].
Notice that [tex]e^{-i\theta}[/tex] has the same modulus that [tex]e^{i\theta}[/tex], so it is on the unit circumference. Beside, its Cartesian representation is [tex](\cos(\theta), -\sin(\theta))[/tex].
So, the points [tex] (\cos(\theta), \sin(\theta))[/tex] and [tex] (\cos(\theta), -\sin(\theta))[/tex] are symmetric with respect to the X-axis. All this can be checked in the attached figure.
(b) Notice that
[tex]e^{i\theta} + e^{-i\theta} = \cos(\theta)+i\sin(\theta) + \cos(\theta)-i\sin(\theta) = 2\cos(\theta)[/tex]
Then,
[tex]\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}[/tex].
(c) Notice that
[tex]e^{i\theta} - e^{-i\theta} = \cos(\theta)+i\sin(\theta) - \cos(\theta)+i\sin(\theta) = 2i\sin(\theta)[/tex]
Then,
[tex]\sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}[/tex].
