Use de moivres theorem to show that tan 3theta=something

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SUMMARY

The discussion focuses on using De Moivre's Theorem to prove the identity for tan 3θ, specifically that tan 3θ = (3tan θ - tan³ θ) / (1 - 3tan² θ). Participants explore the relationship between sine and cosine functions and their tangent equivalents, ultimately suggesting that the problem can be simplified by dividing by cos³ θ. Key identities referenced include tan(α + β) and tan(2α), which are essential for transforming the expressions involved.

PREREQUISITES
  • Understanding of trigonometric identities, specifically tan(α + β) and tan(2α).
  • Familiarity with De Moivre's Theorem and its application in complex numbers.
  • Knowledge of sine and cosine functions and their relationships to tangent.
  • Basic algebraic manipulation skills, particularly in simplifying trigonometric expressions.
NEXT STEPS
  • Study the derivation of tan(3θ) using trigonometric identities.
  • Learn how to apply De Moivre's Theorem in proving trigonometric identities.
  • Explore the relationship between complex numbers and trigonometric functions.
  • Practice simplifying trigonometric expressions by dividing by cosines.
USEFUL FOR

Students studying trigonometry, mathematics educators, and anyone interested in mastering trigonometric identities and their proofs using complex numbers.

blueyellow

Homework Statement




use de moivres theorem to show that
tan 3theta=(3tan theta -tan^3 theta)/(1-3tan^2 theta)

Homework Equations





The Attempt at a Solution


tan 3theta=sin3theta/cos3theta
=(3sin theta - 4 sin^3theta)/(4cos^3theta -3cos theta)

so then i divided everything by cos, and som of the stuff turned into tans, but i can't get it all to turn into tans
 
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blueyellow said:

Homework Statement




use de moivres theorem to show that
tan 3theta=(3tan theta -tan^3 theta)/(1-3tan^2 theta)

Homework Equations





The Attempt at a Solution


tan 3theta=sin3theta/cos3theta
=(3sin theta - 4 sin^3theta)/(4cos^3theta -3cos theta)

so then i divided everything by cos, and som of the stuff turned into tans, but i can't get it all to turn into tans

I don't see how de Moivre's Formula enters into this problem. It seems to me to be a straightforward exercise in proving a trig identity. Although it can be done by converting the tan terms to sine/cosine, it's probably easier to leave things in terms of the tangent.

The identities that you will need are:
tan(\alpha + \beta) = \frac{tan(\alpha) + tan(\beta)}{1 - tan(\alpha)tan(\beta)}
tan(2\alpha) = \frac{2tan(\alpha)}{1 - tan^2(\alpha)}
 
If you really have to use De Moivre's theorem just consider

(cosθ + isinθ)3
 
rock.freak667 said:
If you really have to use De Moivre's theorem just consider

(cosθ + isinθ)3

but that's exactly what i considered and by equating real and imaginary coefficients, i ended up with what i wrote in my first post
and then i don't know how to proceed
any help from anyone would b much appreciated
 
wel, i emailed the lecturer and was told to just divide top and bottom by cos^3 theta
still don't see how that gets rid of the cos's tho
if u hav any idea, pls speak
thanks
 
[Im(cos theta + i sin theta)^3]/[Re(cos theta + i sin theta)^3]

=[3cos^2 theta sin theta - sin^3 theta]/[cos^3 theta - 3 cos theta sin^2 theta]

but how exactly did they go from one step to the next? and how did they get rid of the 'Im' and 'Re'?
 
Does it help that (cos(θ)+ i sin(θ))3 = (cos(3θ)+ i sin(3θ)) ?
 

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