# Math Help - Series expansion?

1. ## Series expansion?

My problem is....

Use the series expansions for sin x and cos x to find the first two
terms of a series expansion for tan x

but which series do i use? Power, maclaurin?
also how do I find tan x ( i know sinx/cosx=tanx) but how do i get there using series?

many thanks

2. Use polynomial division.

3. im having some trouble using polynomial division. what is the method for having a fraction in the nominator/denominator.
when i divide straight down as seen i get

but i know the answer is

if someone could help me with the method please.

4. Originally Posted by decoy808
im having some trouble using polynomial division. what is the method for having a fraction in the nominator/denominator.
when i divide straight down as seen i get

but i know the answer is

if someone could help me with the method please.
You should learn the polynomial long division method. Please refer, Polynomial Long Division. I hope you will find it simple and illustrative.

5. Originally Posted by decoy808
My problem is....

Use the series expansions for sin x and cos x to find the first two
terms of a series expansion for tan x

but which series do i use? Power, maclaurin?
also how do I find tan x ( i know sinx/cosx=tanx) but how do i get there using series?

many thanks
$sinx=x-\frac{x^3}{3!}+\frac{x^5}{5!}-.....$

$cosx=1-\frac{x^2}{2!}+\frac{x^4}{4!}-....$

$tanx=\frac{sinx}{cosx}\Rightarrow\ sinx=cosxtanx$

$\Rightarrow\ x-\frac{x^3}{3!}+\frac{x^5}{5!}-....=\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-....\right)tanx$

Therefore the first term of tanx is x and the 2nd term involves $x^3$ as an $x^2$ would give even powers of x.

$x-\frac{x^3}{3!}+\frac{x^5}{5!}-....=\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-....\right)\left(x+\frac{x^3}{k}+...\right)$

Multiplying out and comparing terms to find k,

$x-\frac{x^3}{3!}+\frac{x^5}{5!}-...=x-\frac{x^3}{2!}+\frac{x^5}{4!}+...+\frac{x^3}{k}-\frac{x^5}{(k)2!}+\frac{x^7}{(k)4!}-...$

$\Rightarrow\frac{ x^3}{k}-\frac{x^3}{2!}=-\frac{x^3}{3!}$

$\Rightarrow\frac{2x^3-kx^3}{(k)2!}=-\frac{x^3}{3!}$

which gives k and therefore the second term in the expansion of tanx.