Prove the following:
sinθ=eiθ−e−iθ2i\sin \theta=\frac {e^{i\theta}-e^{-i\theta}}{2i}sinθ=2ieiθ−e−iθ and cosθ=eiθ+e−iθ2\cos \theta=\frac {e^{i\theta}+e^{-i\theta}}{2}cosθ=2eiθ+e−iθ.
First person to submit an acceptable proof wins my eternal respect:)
Note by A Former Brilliant Member 6 years, 1 month ago
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This follows from Euler's formula which states eiθ=cosθ+isinθe^{i\theta}=\cos\theta +i\sin\thetaeiθ=cosθ+isinθ.
To prove Euler's formula, consider f(t)=eitf(t)=e^{i t}f(t)=eit as a function of time.
Let f(t)f(t)f(t) be the position vector of particle moving in the complex plane at time ttt.
Note that f′(t)=if(t)f'(t)=i f(t)f′(t)=if(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=0t=0, it is equal to 111.
So, the particle is actually moving along a unit circle centered at the origin.
Since ∣f′(t)∣=1|f'(t)|=1∣f′(t)∣=1, we must have f(θ)=eiθ=cosθ+isinθf(\theta)= e^{i\theta}=\cos\theta +i\sin\thetaf(θ)=eiθ=cosθ+isinθ.
We just proved Euler's formula.
Now plug in θ\thetaθ and −θ-\theta−θ 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?
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!}..........ex=1+x+2!x2+3!x3..........
Putting x=iθx=i\thetax=iθ
eiθ=1+iθ−θ22!−iθ33!..........e^{i\theta} = 1+i\theta-\frac{\theta ^2}{2!}-i\frac{\theta^3}{3!}..........eiθ=1+iθ−2!θ2−i3!θ3..........
Grouping terms with iii and without iii
eiθ=(1−θ22!+θ44!......)+i(θ−θ33!+θ55!)......e^{i\theta} = (1-\frac{\theta ^2}{2!}+\frac{\theta^4}{4!}......) +i(\theta-\frac{\theta ^3}{3!} +\frac{\theta ^5}{5!})...... eiθ=(1−2!θ2+4!θ4......)+i(θ−3!θ3+5!θ5)......
eiθ=(cosθ)+i(sinθ)e^{i\theta}= (cos\theta) +i (sin\theta )eiθ=(cosθ)+i(sinθ)
Rest can be done easily as asked by substituting θ\thetaθ and −θ-\theta−θ
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Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
This follows from Euler's formula which states eiθ=cosθ+isinθ.
To prove Euler's formula, consider f(t)=eit 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)=if(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(θ)=eiθ=cosθ+isinθ.
We just proved Euler's formula.
Now plug in θ and −θ and then add (or subtract) to get the formulas you want.
Log in to reply
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?
Mursalin Habib Has provided a very good proof here is another one- ex=1+x+2!x2+3!x3..........
Putting x=iθ
eiθ=1+iθ−2!θ2−i3!θ3..........
Grouping terms with i and without i
eiθ=(1−2!θ2+4!θ4......)+i(θ−3!θ3+5!θ5)......
eiθ=(cosθ)+i(sinθ)
Rest can be done easily as asked by substituting θ and −θ