# Math Help - Maclaurin expansion

1. ## Maclaurin expansion

Find the Maclaurin expansion for $sin^4x$ using the double angle identity for cos2x and cos4x.

So...
This is how i worked out so far, yet i'm not sure whether it is correct or not

Identity of cos4x = cos(2x + 2x) = cos2x.cos2x + sin2x.sin2x
using the identity of $cos2x = (1 - 2sin^2x)(1-2sin^2x) + sin2x.sin2x$

which equals to

$
cos4x = 1 - 4sin^2x + 4sin^4x + sin2x.sin2x
$

hence
$
sin^4x = \frac{cos4x + 4sin^2x - sin2x.sin2x - 1}{4} = sin^4x
$

If it is correct i'm not sure whether to continue with the problem or not, this question has 8 marks...and i'm puzzled.. if anyone could help it would be greatly appreciated!
kleyzam

2. Hello, kleyzam!

Find the Maclaurin expansion for $\sin^4\!x$
using the double-angle identity for $\cos2x$ and $\cos4x.$
I'd approach it like this . . .

$(\sin^2\!x)^2 \;=\;\left(\frac{1-\cos2x}{2}\right)^2 \;=\;\tfrac{1}{4}\left(1 - 2\cos2x + \cos^2\!2x\right)$

. . $= \;\tfrac{1}{4}\left(1 - 2\cos2x + \frac{1+\cos4x}{2}\right) \;=\;\tfrac{1}{8}\left(\cos4x - 4\cos2x + 3\right)$

Then: . $\sin^4\!x \;=\;\frac{1}{8}\begin{Bmatrix}\cos4x \\ -4\cos2x \\ + 3 \end{Bmatrix} \;=$ . $\begin{Bmatrix}1 - \frac{(4x)^2}{2!} + \frac{(4x)^4}{4!} - \frac{(4x)^6}{6!} + \hdots \\ \\[-4mm]
\text{-}4\left[1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \frac{(2x)^6}{6!} + \hdots \right] \\ \\[-4mm] + 3 \end{Bmatrix}$
. . . . etc.