Using e^ix to determine a trig identity

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SUMMARY

The discussion focuses on expressing cosine and sine in terms of the complex exponential function e^ix and its conjugate e^-ix. The identities established are cos(x) = Re(e^ix) and sin(x) = Im(e^ix). The user aims to demonstrate the equation 16cos^3(x)sin^2(x) = 2cos(x) - cos(3x) - cos(5x) using these identities. The conversation emphasizes the utility of Euler's formula, e^(inx) = cos(nx) + i sin(nx), for deriving trigonometric identities.

PREREQUISITES
  • Understanding of Euler's formula: e^(ix) = cos(x) + i sin(x)
  • Familiarity with complex numbers and their properties
  • Knowledge of trigonometric identities and functions
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the derivation of trigonometric identities using Euler's formula
  • Explore the implications of complex exponentials in Fourier analysis
  • Learn about the applications of trigonometric identities in solving differential equations
  • Investigate the relationship between complex numbers and polar coordinates
USEFUL FOR

Students of mathematics, educators teaching trigonometry, and anyone interested in the applications of complex numbers in trigonometric functions.

josephcollins
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Hi people, could someone help me with this

Q. Write cosx and sinx in terms of e^ix and e^-ix respectively

So I wrote that cosx=Re(e^ix)=Re(e^-ix)

and sinx =Im(e^ix) and -Im(e^ix)

I think the above identites are correct, now I must use this to show that

16cos^3(x)sin^2(x) = 2cosx - cos3x - cos5x
 
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[tex]e^{ix} = \cos x + i \sin x[/tex]

[tex]e^{-ix} = \cos x - i \sin x[/tex]

Look at the two above and think how you could rearrange them so you have one for sin(x) in terms of e^(ix) and e^(-ix) and same for cos(x).
 
also remember, [tex]e^{inx} = \cos(nx) + i\sin(nx) = (e^{ix})^n[/tex]. This is all you need to find any trig identity...
 

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