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Complex numbers De Moivres thereom (1 Viewer)

captainvagina

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I hav a question for homework which i just cant seem to figure out

It says Express:
(a) cos4θ in terms of cosθ
(b) sin4θ in terms of cosθ and sinθ
(c) tan4θ in terms of tanθ
 

gurmies

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Keep in mind, when you let tan4@ = sin4@/cos4@, you'll need to divide numerator and denominator by a term that will allow the entire expression to be in terms of tan@.

EDIT: I have forgotten coefficients, lulz.
 
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Dumbledore

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1. cos4@ + i sin4@
= (cos@ + i sin@)^4
= cos^4@ + 4i cos^3@ sin@ - 6cos^2@ sin^2@ -4i cos@ sin^3@ + sin^4@

equating real: cos4@ = cos^4@ - 6cos^2@ sin^2@ + sin^4@
then use sin^2@ = 1-cos^2@ and ur done

2. equate imaginary

3. use sin4@ / cos4@

EDIT: i got beaten xD
 

captainvagina

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thnx heaps guys

yea i got to that line: cos^4@ - 6sin^2@cos^2@ + 4isin@cos^3@ - 4icos@sin^3@ + sin^4@......but didnt know where to go after that. Thnx heaps for the help
 

captainvagina

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ok another question lol.....i equated the real and imaginary parts and im trying to figure out wat tan4@ equals. This is wat i hav im not sure if its right:

tan4@=sin4@/cos4@
tan4@=4isin@cos^3@-4icos@sin^3@/8cos^4@-8cos^2@
tan4@=......dunno where to go lol
 

addikaye03

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I hav a question for homework which i just cant seem to figure out

It says Express:
(a) cos4θ in terms of cosθ
(b) sin4θ in terms of cosθ and sinθ
(c) tan4θ in terms of tanθ
a) (cos@+isin@)^4=cos(4@)+isin(4@)

LHS=c^4+4c^3(is)+6c^2(is)^2+4c(is)^3+(is)^4

=cos^4@+i4cos^3@sin@-6cos^2@sin^2@-i4cos@sin^3@+sin^4@

Taking real

cos(4@)=cos^4@-6cos^2@sin^2@+sin^4@ where sin^2@=1-cos^2@

b) Equate imaginary:

c) sin(4@)/cos(4@)


EDIT: wow, really gotta refresh page after it's been sitting here lol beaten about 30 times lol
 
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Dumbledore

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ok another question lol.....i equated the real and imaginary parts and im trying to figure out wat tan4@ equals. This is wat i hav im not sure if its right:

tan4@=sin4@/cos4@
tan4@=4isin@cos^3@-4icos@sin^3@/8cos^4@-8cos^2@
tan4@=......dunno where to go lol
divide both the numerator and denominator by cos^4 @ the cos' will cancel and sins will turn into tans if u get sec then u can use 1+tan^2 = sec^2

btw there is no "i" in there cause ur equating complex parts
 

ninetypercent

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lol yeah. it makes it seem as though you copied from the person above
 

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