Euler

[logistic_guy]

Using Euler's relation, derive the following relationships:

(a) \(\displaystyle \cos \theta = \frac{1}{2}(e^{j\theta} + e^{-j\theta})\)

(b) \(\displaystyle \sin \theta = \frac{1}{2j}(e^{j\theta} - e^{-j\theta})\)

(c) \(\displaystyle \cos^2 \theta = \frac{1}{2}(1 + \cos 2\theta)\)

(d) \(\displaystyle \sin \theta \sin \phi = \frac{1}{2}\cos(\theta - \phi) - \frac{1}{2}\cos(\theta + \phi)\)

(e) \(\displaystyle \sin(\theta + \phi) = \sin \theta \cos \phi + \cos \theta \sin \phi\)

 

[khansaheb]

Using Euler's relation, derive the following relationships:

(a) \(\displaystyle \cos \theta = \frac{1}{2}(e^{j\theta} + e^{-j\theta})\)

(b) \(\displaystyle \sin \theta = \frac{1}{2j}(e^{j\theta} - e^{-j\theta})\)

(c) \(\displaystyle \cos^2 \theta = \frac{1}{2}(1 + \cos 2\theta)\)

(d) \(\displaystyle \sin \theta \sin \phi = \frac{1}{2}\cos(\theta - \phi) - \frac{1}{2}\cos(\theta + \phi)\)

(e) \(\displaystyle \sin(\theta + \phi) = \sin \theta \cos \phi + \cos \theta \sin \phi\)

show us your effort/s to solve this problem.