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.
$\displaystyle (cosx+isinx)^5=cos5x+isin5x$
Expanding
$\displaystyle cos^5x+i5cos^4x{sinx}-10cos^3x{sin^2x}-i10cos^2x{sin^3x}+5cosx{sin^4x}+isin^5x$
$\displaystyle =cos5x+isin5x$
Now compare terms with real coefficients, these terms are those that will make $\displaystyle cos5x$
Hello, sf1903!
Use de Moivre's formula and the Binomial theorem to derive a formula
that expresses $\displaystyle \cos5x$ as a polynomial in $\displaystyle \cos x$
DeMoivre's formula says: .$\displaystyle \cos5x + i\sin5x \:=\:(\cos x + i\sin x)^5$ .[1]
Expand the right side:
. .$\displaystyle \cos5x + i\sin5x \:=\:(\cos x + i\sin x)^5$
. . . . $\displaystyle =\; \cos^5\!x + 5i\cos^4\!x\sin x - 10\cos^3\!x\sin^2\!x - 10i\cos^2\!x\sin^3\!x + 5\cos x\sin^4\!x + i\sin^5\!x$
. . . . $\displaystyle =\; \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:
$\displaystyle \cos5x \:=\: \cos^5\!x - 10\cos^3\!x\sin^2\!x + 5\cos x\sin^4\!x $
. . . . $\displaystyle =\;\cos^5\!x - 10\cos^3\!x(1-\cos^2\!x) + 5\cos x(1-\cos^2\!x)^2 $
. . . . $\displaystyle = \;\cos^5\!x - 10\cos^3\!x + 10\cos^5\!x + 5\cos x - 10\cos^3\!x + 5\cos^5\!x $
$\displaystyle \boxed{\cos5x \:=\:16\cos^5\!x - 20\cos^3\!x + 5\cos x} $
$\displaystyle \sin5x \:=\:5\cos^4\!x\sin x - 10\cos^2\!x\sin^3\!x + \sin^5\!x$
. . . . $\displaystyle =\; 5(1-\sin^2\!x)^2\sin x - 10(1-\sin^2\!x)\sin^3\!x + \sin^5\!x$
. . . . $\displaystyle =\; 5\sin x - 10\sin^3\!x + 5\sin^5\!x - 10\sin^3\!x + 10\sin^5\!x + \sin^5\!x $
$\displaystyle \boxed{\sin 5x \:=\:5\sin x - 20\sin^3\!x + 16\sin^5\!x} $