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**1. Simplify e**

^{i6x}(1+e^{-i10x})/(1+e^{i2x})**2. i have no idea how to simplify this its supposed to be in terms of cosines**

**3. i dont how i can simplify this such that i can use the 1/2(e^x +e^-x) = cosx formula**

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- Thread starter Luongo
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- #2

fzero

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You need the Euler formula [tex]e^{ix} = \cos x + i \sin x[/tex].

- #3

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You need the Euler formula [tex]e^{ix} = \cos x + i \sin x[/tex].

i tried that, i got garbage it MUST be in terms of cosines

- #4

fzero

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well, ive multiplied it out i got e^6x+e^-4x and then i did e^2x * e^4x + e^-4x

in the numerator to try to get it in cosine form but i cant get the e^2 out of there so i really have no idea how, can someone show me the steps of simplifying this

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fzero

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i got it! you're a genius, how on earth did you see that?!?!

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Exponent rules :D...

- #9

fzero

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i got it! you're a genius, how on earth did you see that?!?!

You know that

[tex]\frac{ e^{ia} + e^{-ia}}{2} = \cos a[/tex]

so if you see

[tex] 1 + e^{ib}[/tex]

you want to rewrite that as

[tex] e^{ib/2} ( e^{-ib/2} + e^{ib/2} ).[/tex]

- #10

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You know that

[tex]\frac{ e^{ia} + e^{-ia}}{2} = \cos a[/tex]

so if you see

[tex] 1 + e^{ib}[/tex]

you want to rewrite that as

[tex] e^{ib/2} ( e^{-ib/2} + e^{ib/2} ).[/tex]

thanks alot i appreciate it i was hung up on this question for a while

- #11

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thanks alot i appreciate it i was hung up on this question for a while

for simplifying in terms of sines... can i use the same formula except the negative sign is between the 2 expos?

- #12

fzero

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for simplifying in terms of sines... can i use the same formula except the negative sign is between the 2 expos?

Yes.

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