Prove the following:

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

First person to submit an acceptable proof wins my eternal respect:)

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## Comments

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TopNewestThis follows from Euler's formula which states \(e^{i\theta}=\cos\theta +i\sin\theta\).

To prove Euler's formula, consider \(f(t)=e^{i t}\) as a function of time.

Let \(f(t)\) be the position vector of particle moving in the complex plane at time \(t\).

Note that \(f'(t)=i f(t)\). This means the velocity of the particle is always perpendicular to its position vector. This means that the magnitude of the position vector never changes. At \(t=0\), it is equal to \(1\).

So, the particle is actually moving along a unit circle centered at the origin.

Since \(|f'(t)|=1\), we must have \(f(\theta)= e^{i\theta}=\cos\theta +i\sin\theta\).

We just proved Euler's formula.

Now plug in \(\theta\) and \(-\theta\) and then add (or subtract) to get the formulas you want.

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I've never seen that proof before - only the Taylor series one. Also, where you have said "This means that the velocity of the particle is always perpendicular to its position vector" doesn't that imply that the tangent to a radius circle theorem is coming into play?

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Mursalin Habib Has provided a very good proof here is another one- \(e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}..........\)

Putting \(x=i\theta\)

\(e^{i\theta} = 1+i\theta-\frac{\theta ^2}{2!}-i\frac{\theta^3}{3!}..........\)

Grouping terms with \(i\) and without \(i\)

\(e^{i\theta} = (1-\frac{\theta ^2}{2!}+\frac{\theta^4}{4!}......) +i(\theta-\frac{\theta ^3}{3!} +\frac{\theta ^5}{5!})...... \)

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

Rest can be done easily as asked by substituting \(\theta\) and \(-\theta\)

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