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Crazy boy [7]
3 years ago
11

Prove Euler's identity using Euler's formula. e^ix = cos x + i sin x

Mathematics
1 answer:
Korolek [52]3 years ago
3 0

First list all the terms out.

e^ix = 1 + ix/1! + (ix)^2/2! + (ix)^3/3! ...

Then, we can expand them.

e^ix = 1 + ix/1! + i^2x^2/2! + i^3x^3/3!...

Then, we can use the rules of raising i to a power.

e^ix = 1 + ix - x^2/2! - ix^3/3!...

Then, we can sort all the real and imaginary terms.

e^ix = (1 - x^2/2!...) + i(x - x^3/3!...)

We can simplify this.

e^ix = cos x + i sin x

This is Euler's Formula.

What happens if we put in pi?

x = pi

e^i*pi = cos(pi) + i sin(pi)

cos(pi) = -1

i sin(pi) = 0

e^i*pi = -1 OR e^i*pi + 1 = 0

That is Euler's identity.

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Step 1:
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\frac{d}{dx}[y cos x]= \frac{d}{dx}[5x^2]+ \frac{d}{dx}[ 3y^2]
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Step 2:

Deal with the terms in 'x' and the constant terms
\frac{d}{dx}[ycosx]= 10x+ \frac{d}{dx} [3y^2]
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Step 3:

Use the chain rule for the terms in 'y'
\frac{d}{dx}[ycosx]=10x+6y \frac{dy}{dx}
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Step 4:

Use the product rule on the term in 'x' and 'y'
(y) \frac{d}{dx} cos x+(cos x) \frac{d}{dx}y =10x+6y \frac{dy}{dx}

y(-siny)+(cosx) \frac{dy}{dx} =10x+6y \frac{dy}{dx}
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Step 5:

Rearrange to make \frac{dy}{dx} the subject
-y sin(y)+cos(x) \frac{dy}{dx} =10x+6y \frac{dy}{dx}
cos(x)  \frac{dy}{dx}-6y \frac{dy}{dx}=10x+y sin(y)
[cos(x) - 6y]  \frac{dy}{dx}=10x + y sin(y)
\frac{dy}{dx}= \frac{10x+ysin(y)}{cos(x)-6y} ⇒ Final Answer


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