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.

 

[logistic_guy]

show us your effort/s to solve this problem.

I will start with Euler's relation: \(\displaystyle e^{j\theta} = \cos \theta + j\sin \theta\).

(a)

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

(b)

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

(c)

\(\displaystyle \cos^2 \theta = \cos \theta \cos \theta = \frac{1}{2}\left(e^{j\theta} + e^{-j\theta}\right)\frac{1}{2}\left(e^{j\theta} + e^{-j\theta}\right) = \frac{1}{4}\left(e^{j\theta + j\theta} + e^{j\theta - j\theta} + e^{-j\theta + j\theta} + e^{-j\theta - j\theta}\right)\)
\(\displaystyle = \frac{1}{4}\left(e^{2j\theta} + 2e^{0} + e^{-2j\theta}\right) = \frac{1}{2}\left[1 + \frac{1}{2}\left(e^{2j\theta} + e^{-2j\theta}\right)\right] = \frac{1}{2}\left(1 + \cos 2\theta\right)\)

I will continue in the next post.

 

[fresh_42]

See, it is all about what can be used to prove something. The prerequisites are crucial.

Do you have any idea how to prove (c)-(e) by geometric means?

 

[logistic_guy]

See, it is all about what can be used to prove something. The prerequisites are crucial.

Do you have any idea how to prove (c)-(e) by geometric means?

Thanks professor fresh_42 for being interested in my posts

Let me try it for (e). Here is some sketch I made after a few tries:



\(\displaystyle \text{BC} = \cos y \)
\(\displaystyle \text{BF} = \sin y \)
\(\displaystyle \text{FG} = \cos x \times \text{BF} = \cos x \sin y \)
\(\displaystyle \text{AB} = \sin x \times \text{BC} = \sin x \cos y \)
\(\displaystyle \sin(x + y) = \text{DF} = \text{DG} \ + \text{FG} = \text{AB} \ + \text{FG} = \sin x \cos y + \cos x \sin y\)

 

[logistic_guy]

Let me try to prove (c) this time. I made this sketch:



Let me first try to find the angles, \(\displaystyle \angle n\) and \(\displaystyle \angle y\).

\(\displaystyle \angle n = 180^{\circ} - 90^{\circ} - x = 90^{\circ} - x\).

\(\displaystyle \angle n + \angle n + \angle y = 180^{\circ}\)

\(\displaystyle 2\angle n + \angle y = 180^{\circ}\)

\(\displaystyle 2(90^{\circ} - x) + \angle y = 180^{\circ}\)

\(\displaystyle 180^{\circ} - 2x + \angle y = 180^{\circ}\)

\(\displaystyle \angle y = 2x\)

\(\displaystyle \text{EF} = \cos y = \cos 2x\)

\(\displaystyle \Delta \text{ABC} \sim \Delta AFB\)

This means \(\displaystyle \frac{\text{AB}}{\text{AC}} = \frac{AF}{AB} = \frac{1 + \text{EF}}{AB}\)

\(\displaystyle \frac{2\cos x}{2} = \frac{1 + \cos 2x}{2\cos x}\)

\(\displaystyle 2\cos^2 x = 1 + \cos 2x\)

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