# Math Help - De Moivre's formula and Binomial theorem

1. ## De Moivre's formula and Binomial theorem

Use de Moivre's formula and the Binomial theorem to derive a formula that expresses cos5x as a polynomial in cosx, or more precisely determine the constants ABC.

This is what I have done.. is it correct or im I totally wrong?

2. The question says to use De' Moivre's Theorem and I didn't see you use it (I could be wrong), so I'll start it.

$(cosx+isinx)^5=cos5x+isin5x$
Expanding
$cos^5x+i5cos^4x{sinx}-10cos^3x{sin^2x}-i10cos^2x{sin^3x}+5cosx{sin^4x}+isin^5x$
$=cos5x+isin5x$

Now compare terms with real coefficients, these terms are those that will make $cos5x$

3. Originally Posted by I-Think
The question says to use De' Moivre's Theorem and I didn't see you use it (I could be wrong), so I'll start it.

$(cosx+isinx)^5=cos5x+isin5x$
Expanding
$cos^5x+i5cos^4x{sinx}-10cos^3x{sin^2x}-i10cos^2x{sin^3x}+5cosx{sin^4x}+isin^5x$
$=cos5x+isin5x$

Now compare terms with real coefficients, these terms are those that will make $cos5x$

Do you mean that A = 1, B = -10, C= 5 ???

4. Hello, sf1903!

Use de Moivre's formula and the Binomial theorem to derive a formula
that expresses $\cos5x$ as a polynomial in $\cos x$

DeMoivre's formula says: . $\cos5x + i\sin5x \:=\:(\cos x + i\sin x)^5$ .[1]

Expand the right side:

. . $\cos5x + i\sin5x \:=\:(\cos x + i\sin x)^5$

. . . . $=\; \cos^5\!x + 5i\cos^4\!x\sin x - 10\cos^3\!x\sin^2\!x - 10i\cos^2\!x\sin^3\!x + 5\cos x\sin^4\!x + i\sin^5\!x$

. . . . $=\; \left(\cos^5\!x -10\cos^3\!x\sin^2\!x + 5\cos x\sin^4\!x\right) + i\left(5\cos^4\!x\sin x - 10\cos^2\!x\sin^3\!x + \sin^5\!x\right)$

Equate real and imaginary components:

$\cos5x \:=\: \cos^5\!x - 10\cos^3\!x\sin^2\!x + 5\cos x\sin^4\!x$

. . . . $=\;\cos^5\!x - 10\cos^3\!x(1-\cos^2\!x) + 5\cos x(1-\cos^2\!x)^2$

. . . . $= \;\cos^5\!x - 10\cos^3\!x + 10\cos^5\!x + 5\cos x - 10\cos^3\!x + 5\cos^5\!x$

$\boxed{\cos5x \:=\:16\cos^5\!x - 20\cos^3\!x + 5\cos x}$

$\sin5x \:=\:5\cos^4\!x\sin x - 10\cos^2\!x\sin^3\!x + \sin^5\!x$

. . . . $=\; 5(1-\sin^2\!x)^2\sin x - 10(1-\sin^2\!x)\sin^3\!x + \sin^5\!x$

. . . . $=\; 5\sin x - 10\sin^3\!x + 5\sin^5\!x - 10\sin^3\!x + 10\sin^5\!x + \sin^5\!x$

$\boxed{\sin 5x \:=\:5\sin x - 20\sin^3\!x + 16\sin^5\!x}$