Proofs: Trigonometrical!

Prove the following:

sinθ=eiθeiθ2i\sin \theta=\frac {e^{i\theta}-e^{-i\theta}}{2i} and cosθ=eiθ+eiθ2\cos \theta=\frac {e^{i\theta}+e^{-i\theta}}{2}.

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

#Algebra #Trigonometry #Sine #Cosine #Theta

Note by A Former Brilliant Member
6 years, 1 month ago

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Comments

This follows from Euler's formula which states eiθ=cosθ+isinθe^{i\theta}=\cos\theta +i\sin\theta.

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

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

Note that f(t)=if(t)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=0t=0, it is equal to 11.

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

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

We just proved Euler's formula.

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

Mursalin Habib - 6 years, 1 month ago

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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?

Curtis Clement - 6 years, 1 month ago

Mursalin Habib Has provided a very good proof here is another one- ex=1+x+x22!+x33!..........e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}..........

Putting x=iθx=i\theta

eiθ=1+iθθ22!iθ33!..........e^{i\theta} = 1+i\theta-\frac{\theta ^2}{2!}-i\frac{\theta^3}{3!}..........

Grouping terms with ii and without ii

eiθ=(1θ22!+θ44!......)+i(θθ33!+θ55!)......e^{i\theta} = (1-\frac{\theta ^2}{2!}+\frac{\theta^4}{4!}......) +i(\theta-\frac{\theta ^3}{3!} +\frac{\theta ^5}{5!})......

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

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

Gautam Sharma - 6 years, 1 month ago
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