Complex Numbers - De Moivres Theorem (1 Viewer)

[mazza_728]

Hey.. just doing some revision and i need to express cons2@-isin2@ in the form (cos@+isin@)ⁿ and also need to show, using de moivre's theorem that
cos2@=cos²@-isin²@
sin2@=2sin@cos@
and
tan2@=(2tan@)/(1-tan²@)
Can you guys help??
Thanks xoxo

 

[mojako]

expand (cos@+isin@)n (n=2) using De Moivre's theorem.
expand (cos@+isin@)n (n=2) using binomial theorem.

equate real parts of the two expansions
equate imaginary parts of the two expansions

tan2@=(2tan@)/(1-tan2@)
for this one I think you can do sin2@/cos2@
(write sin2@=2sin@cos@ etc)

 

[BillyMak]

Hey.. just doing some revision and i need to express cons2@-isin2@ in the form (cos@+isin@)ⁿ and also need to show, using de moivre's theorem that
cos2@=cos²@-isin²@
sin2@=2sin@cos@
and
tan2@=(2tan@)/(1-tan²@)
Can you guys help??
Thanks xoxo

cos2@ - isin2@ = (cos@ - isin@)²

RHS = (cos@ - isin@)²
= cos²@ - 2isin@cos@ - sin²@

Equate Real parts from LHS and RHS:

cos2@ = cos²@ - sin²@ - Eqn. 1 (That "i" you gave in the question shouldn't be there methinks)

Equate Imaginary parts from LHS and RHS:

- sin2@ = - 2sin@cos@
sin2@ = 2sin@cos@ - Eqn. 2

Eqn.2/Eqn.1 = (sin2@)/(cos2@) = 2sin@cos@/(cos²@ - sin²@)

Divide top and bottom of RHS by cos²@

tan2@ = 2tan@/(1 - tan²@)

 

[ngai]

Hey.. just doing some revision and i need to express cons2@-isin2@ in the form (cos@+isin@)ⁿ

cons2@
how did u manage to type cons?
the n is nowhere near the c, o, s, or @

 

[BillyMak]

he's used to writing cons2pated! HA! HA!!



HA!













ha...

 

[Teoh]

Hehehe

Force of habit? Most words that start with 'con' end up having the next letter as 's'?

Conscription? Consumption? Conspicuous? Constituition?

We can make a game of it

 

[Constip8edSkunk]

he's used to writing cons2pated! HA! HA!!



HA!













ha...

consTIpated ... not consTOpated