Use the binomial expansion to find the real and imaginary parts of $(cosθ+isinθ)^5$ Hence show that $sin5θ/sinθ=16cos^4θ-12cos^2θ+1$

I expanded this expression and I got: $cos^5θ+5icos^4θsinθ-10cos^3θsin^2θ-10icos^2θsin^3θ+5cosθsin^4θ+isin^5θ$

Then I used the Moivre's theorem and I got: $(cos5θ+isin5θ)$

I compared the imaginary parts and I got something like: $sin5θ=5cos^4θsinθ-10cos^2θsin^3θ+sin^5θ$

which is very close to: $(16cos^4θ-12cos^2θ+1)sinθ$ but not the same.

Where do I make te mistake?

Thanks for any help! ;)

  • 2
    $\begingroup$ Have you tried use $\cos^2\theta+sin^2\theta =1$ $\endgroup$
    – Fan
    May 15 '17 at 19:13
  • 1
    $\begingroup$ You didn't err. You just didn't finish. $\endgroup$ May 15 '17 at 19:35


In your last line, factor by $\sin (\theta), $


$\sin^2(\theta) $ by $1-\cos^2 (\theta)$ and

$\sin^4 (\theta) $ by

$(1-\cos^2 (\theta))^2=1+\cos^4 (\theta)-2\cos^2 (\theta) $

you will get it.


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