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Use the Euler identity to prove sin^2x+cos^2x=1

1. The problem statement, all variables and given/known data

Just like my title says, we are to prove the trig identity sin^2x+cos^2x=1 using the Euler identity.

2. Relevant equations

Euler - e^(ix) = cosx + isinx
trig identity - sin^2x + cos^2x = 1

3. The attempt at a solution

I tried solving the Euler for sinx and cosx, then plugging it into the trig identity. I just ended up with something really sloppy with multiple i's left over. I also tried squaring Euler and manipulating values that way to get an answer, but that just ended up really sloppy as well.

If anyone has any tips, I'd really appreciate it. I spent about 30mins on this now and am just at a loss.
 

HallsofIvy

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I presume, then, that you got [tex]sin(x)= \frac{e^{ix}- e^{-ix}}{2i}[/tex]
and [tex]cos(x)= \frac{e^{ix}+ e^{-ix}}{2}[/tex]

So that [tex]sin^2(x)= \frac{e^{2ix}- 2+ e^{-2ix}}{-4}[/tex]
and [tex]cos^2(x)= \frac{e^{2ix}+ 2+ e^{-2ix}}{4}[/tex]

That doesn't look very "sloppy" to me. What happens when you add them? (Be careful of that "-" in the denominator of [itex]sin^2(x)[/itex].)
 
I presume, then, that you got [tex]sin(x)= \frac{e^{ix}- e^{-ix}}{2i}[/tex]
and [tex]cos(x)= \frac{e^{ix}+ e^{-ix}}{2}[/tex]

So that [tex]sin^2(x)= \frac{e^{2ix}- 2+ e^{-2ix}}{-4}[/tex]
and [tex]cos^2(x)= \frac{e^{2ix}+ 2+ e^{-2ix}}{4}[/tex]

That doesn't look very "sloppy" to me. What happens when you add them? (Be careful of that "-" in the denominator of [itex]sin^2(x)[/itex].)
Hmm, that is much different than what I got.

When I solved with Euler for cosx, I got:

cosx = e^(ix)-isinx, then I squared that and that is when it got sloppy.

I'm not sure how you got rid of the "isinx" and got the rest of what you did in your equations?
 

Orodruin

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1. The problem statement, all variables and given/known data

Just like my title says, we are to prove the trig identity sin^2x+cos^2x=1 using the Euler identity.

2. Relevant equations

Euler - e^(ix) = cosx + isinx
trig identity - sin^2x + cos^2x = 1

3. The attempt at a solution

I tried solving the Euler for sinx and cosx, then plugging it into the trig identity. I just ended up with something really sloppy with multiple i's left over. I also tried squaring Euler and manipulating values that way to get an answer, but that just ended up really sloppy as well.

If anyone has any tips, I'd really appreciate it. I spent about 30mins on this now and am just at a loss.
I suggest an alternative route. If
$$
e^{ix} = \cos x + i \sin x
$$
what is then ##e^{-ix}##?
 
I suggest an alternative route. If
$$
e^{ix} = \cos x + i \sin x
$$
what is then ##e^{-ix}##?
I'm sorry, but I'm not sure where that is going? Do you mean just 1/(cosx +isinx)? I don't see what to do with that? My problem is I keep ending up with "i" somewhere in my equations that I can't get rid of.
 

HallsofIvy

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No, he means replace "x" with "-x". If [itex]e^{ix}= cos(x)+ isin(x)[/itex] then [itex]e^{-ix}= cos(-x)+ isin(-x)= cos(x)- isin(x)[/itex]. Now you can solve for cos(x) in terms of [itex]e^{ix}[/itex] and [itex]e^{-ix}[/tex] by adding those two equation and solve for sin(x) by subtracting.
 

olivermsun

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He means if ##e^{ix} = \cos x + i \sin x,##

then ##e^{-ix} = \cos (-x) + i \sin (-x) = \cos (x) - i \sin (x).##

Then you have enough pieces to do some creative assembly and get the identity in question.

Edit: Never mind, Halls just posted.
 
Last edited:
Oh, thank you guys. I didn't know I was able to do that. Clearly I have some things to learn this semester. Now it's a piece of cake. I appreciate the help.
 

Orodruin

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Now you can solve for cos(x) in terms of [itex]e^{ix}[/itex] and [itex]e^{-ix}[/tex] by adding those two equation and solve for sin(x) by subtracting.
No, I intended for him to multiply them. Solving for the cos and sin seems like an unnecessary detour.
 

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