Question

Can anybody point me to online resources that talk about using de moivre's theorem to prove trig identities like below:
sin 2x = 2 cos x. sin x

Answer 1

http://openstudy.com/users/watchmath#/users/watchmath/updates/4dd7c19dd95c8b0bdd3a61c4

Answer 2

hmm. the idea is this. \[e^{i\theta}=cos(\theta)+isin(\theta)\] and
\[(e^i\theta)^2=e^{2i\theta}=(cos(\theta)+isin(\theta))^2=cos({2\theta})+isin(2\theta)\]

Answer 3

square \[(cos(\theta) + i sin(\theta))\] to see what you get. then equate the real part to the real part and you get an identity for \[cos(2\theta)\] and another one for \[sin(2\theta)\]

Answer 4

http://math.mit.edu/classes/18.013A/MathML/chapter18/section04.xhtml

Answer 5

when you compute
\[(cos(\theta)+isin(\theta))^2\]
you get \[cos^2(\theta)-sin^2(\theta)+i\times 2 cos(\theta)sin(\theta)\]
the real part is \[cos^2(\theta)-sin^2(\theta)\] so that must equal the real part of \[cos(2\theta)+isin(2\theta)\] which is just \[cos(2\theta)\] telling you that
\[cosd(2\theta)=cos^2(\theta)-sin^2(\theta)\]

Answer 6

likewise
\[sin(2\theta)=2cos(\theta)sin(\theta)\]

Answer 7

My concern is: what de moivre's theorem says?

z^n = r^n (( cos nx) + i sin (n x)) or
(cos x + i sin x) ^n= ( cos nx) + i sin (n x)

I am confused.

Answer 8

both are true.

Answer 9

How come> can u explain?

Answer 10

\[z=re^{i\theta}=r(cos(\theta)+isin(\theta))\]

Answer 11

\[z^n=r^n(e^{i\theta})^n=r^ne^{ni\theta}\]

Answer 12

that by the laws of exponents.

Answer 13

and since \[e^{ni\theta}=cos(n\theta)+isin(n\theta) \] you get the second equality

Answer 14

if you have not seen
\[z=re^{i\theta}\] as a representation of a complex number, then it requires a difffernt explanation, but if you have seen it it is nothing more than the laws of exponents.

Answer 15

succinctly put here
http://en.wikipedia.org/wiki/De_Moivre%27s_formula

Answer 16

Thanks .I need to site on this, it has been itching my head since yesterday.I appreciate your help.

Answer 17

welcome
hope at least second explanation was clear.

Answer 18

i must have made another algebra error let me check.

Answer 19

if you have not seen
\[z=re^{i\theta}\] as a representation of a complex number, then it requires a difffernt explanation, but if you have seen it it is nothing more than the laws of exponents.