# sin^4 θ as linear combo of e^ikθ

• Oct 23rd 2010, 08:25 AM
SyNtHeSiS
sin^4 θ as linear combo of e^ikθ
Use the Binomial Theorem to write $\displaystyle sin^4\theta$ as a linear combo of functions of the form $\displaystyle e^{ik\theta}$, where k E Z.

Attempt:

$\displaystyle sin^4 \theta$
$\displaystyle = (\frac{e^{i\theta} - e^{-i \theta}}{2i})^4$
$\displaystyle = \frac{1}{16}(e^{i\theta} - e^{-i\theta})^4$

$\displaystyle (e^{i\theta} - e^{-i\theta})^4 = \sum_{k=0}^{4}\binom{4}{k}$$\displaystyle (e^{i\theta})^{4 - k}(-e^{-i\theta})^k \displaystyle =\binom{4}{0}(e^{i\theta})^4 + \binom{4}{1}(e^{i\theta})^3(-e^{-i\theta}) + \binom{4}{2}(e^{i\theta})^2(-e^{-i\theta})^2 + \binom{4}{3}(e^{i\theta})(-e^{-i\theta})^3 + (-e^{-i\theta})^4 \displaystyle =e^{i4\theta} - 4e^{i3\theta - i\theta} + 6e^{i2\theta + i2\theta} - 4e^{i\theta - i3\theta} + e^{i4\theta} \displaystyle =e^{i4\theta} - 4e^{i2\theta} + 6e^{i4\theta} - 4e^{-i2\theta} + e^{i4\theta} \displaystyle =8e^{i4\theta} - 4e^{i2\theta} - 4e^{-i2\theta} \displaystyle sin^4 \theta \displaystyle = \frac{1}{16}(8e^{i4\theta} - 4e^{i2\theta} - 4e^{-i2\theta}) \displaystyle =\frac{1}{2}e^{4\theta} - \frac{1}{4}e^{i2\theta} - \frac{1}{4}e^{-{i2\theta}} but I think I made a mistake somewhere • Oct 23rd 2010, 09:30 AM Plato It should be just \displaystyle 6 not \displaystyle 6e^{4\theta i}. • Oct 23rd 2010, 09:30 AM Jhevon Quote: Originally Posted by SyNtHeSiS Use the Binomial Theorem to write \displaystyle sin^4\theta as a linear combo of functions of the form \displaystyle e^{ik\theta}, where k E Z. Attempt: \displaystyle sin^4 \theta \displaystyle = (\frac{e^{i\theta} - e^{-i \theta}}{2i})^4 \displaystyle = \frac{1}{16}(e^{i\theta} - e^{-i\theta})^4 \displaystyle (e^{i\theta} - e^{-i\theta})^4 = \sum_{k=0}^{4}\binom{4}{k}$$\displaystyle (e^{i\theta})^{4 - k}(-e^{-i\theta})^k$
$\displaystyle =\binom{4}{0}(e^{i\theta})^4 + \binom{4}{1}(e^{i\theta})^3(-e^{-i\theta}) + \binom{4}{2}(e^{i\theta})^2(-e^{-i\theta})^2 + \binom{4}{3}(e^{i\theta})(-e^{-i\theta})^3 + (-e^{-i\theta})^4$
$\displaystyle =e^{i4\theta} - 4e^{i3\theta - i\theta} + 6e^{i2\theta + i2\theta} - 4e^{i\theta - i3\theta} + e^{i4\theta}$
$\displaystyle =e^{i4\theta} - 4e^{i2\theta} + 6e^{i4\theta} - 4e^{-i2\theta} + e^{i4\theta}$
$\displaystyle =8e^{i4\theta} - 4e^{i2\theta} - 4e^{-i2\theta}$

$\displaystyle sin^4 \theta$
$\displaystyle = \frac{1}{16}(8e^{i4\theta} - 4e^{i2\theta} - 4e^{-i2\theta})$
$\displaystyle =\frac{1}{2}e^{4\theta} - \frac{1}{4}e^{i2\theta} - \frac{1}{4}e^{-{i2\theta}}$

but I think I made a mistake somewhere

you did. note that $\displaystyle (-e^{-x})^2 = e^{-2x}$ and similarly, $\displaystyle (-e^{-x})^4 = e^{-4x}$

EDIT: which explains Plato's comment
• Oct 24th 2010, 03:09 AM
SyNtHeSiS
Thanks. So would the correct answer be:

$\displaystyle sin^4\theta = \frac{1}{8}cos4\theta = \frac{1}{2}cos2\theta + \frac{3}{8}$?
• Oct 24th 2010, 08:18 AM
Jhevon
Quote:

Originally Posted by SyNtHeSiS
Thanks. So would the correct answer be:

$\displaystyle sin^4\theta = \frac{1}{8}cos4\theta = \frac{1}{2}cos2\theta + \frac{3}{8}$?

??

• Oct 25th 2010, 04:41 AM
SyNtHeSiS
Quote:

Originally Posted by Jhevon

Yeah I did. Here is my working out:

$\displaystyle sin^4 \theta$
$\displaystyle = (\frac{e^{i\theta} - e^{-i \theta}}{2i})^4$
$\displaystyle = \frac{1}{16}(e^{i\theta} - e^{-i\theta})^4$

$\displaystyle (e^{i\theta} - e^{-i\theta})^4 = \sum_{k=0}^{4}\binom{4}{k}$$\displaystyle (e^{i\theta})^{4 - k}(-e^{-i\theta})^k$
$\displaystyle =\binom{4}{0}(e^{i\theta})^4 + \binom{4}{1}(e^{i\theta})^3(-e^{-i\theta}) + \binom{4}{2}(e^{i\theta})^2(-e^{-i\theta})^2 + \binom{4}{3}(e^{i\theta})(-e^{-i\theta})^3 + (-e^{-i\theta})^4$
$\displaystyle =e^{i4\theta} - 4e^{i3\theta - i\theta} + 6e^{i2\theta - i2\theta} - 4e^{i\theta - i3\theta} + e^{-i4\theta}$
$\displaystyle =e^{i4\theta} - 4e^{i2\theta} + 6 - 4e^{-i2\theta} + e^{-i4\theta}$

$\displaystyle sin^4 \theta$
$\displaystyle =\frac{1}{16}(e^{i4\theta} - 4e^{i2\theta} + 6 - 4e^{-i2\theta} + e^{-i4\theta})$
$\displaystyle =\frac{1}{8}(\frac{e^{i4\theta + e^{-i4\theta}}}{2}) - \frac{1}{2}(\frac{e^{i2\theta} + e^{-i2\theta}}{2}) + \frac{3}{8}$
$\displaystyle =\frac{1}{8}cos4\theta = \frac{1}{2}cos2\theta + \frac{3}{8}$