# prove tanx =

• June 23rd 2009, 11:31 PM
prove tanx =
hi

I'm trying to prove: tanx = x + (1/3)x^3 + (2/15)x^5 +...

using sinx = x - (1/3!)x^3 + (1/5!)x^5 -...

and cosx = 1 - (1/2!)x^2 + (1/4!)x^4 -...

my working is as follows:

tanx=sinx/cosx = (sinx)(cosx)^-1

I then take

(cosx)^-1 = [1 - (1/2!)x^2 + (1/4!)x^4 -...]^-1 ......(i)

and try using the binomial theorem

(1 + x)^-1 = 1 -x + x^2 - x^3 +... .......(ii)

but when i try to substitute (i) into (ii) i come unstuck!!

• June 24th 2009, 02:08 AM
Quote:

hi

I'm trying to prove: tanx = x + (1/3)x^3 + (2/15)x^5 +...

using sinx = x - (1/3!)x^3 + (1/5!)x^5 -...

and cosx = 1 - (1/2!)x^2 + (1/4!)x^4 -...

my working is as follows:

tanx=sinx/cosx = (sinx)(cosx)^-1

I then take

(cosx)^-1 = [1 - (1/2!)x^2 + (1/4!)x^4 -...]^-1 ......(i)

and try using the binomial theorem

(1 + x)^-1 = 1 -x + x^2 - x^3 +... .......(ii)

but when i try to substitute (i) into (ii) i come unstuck!!

Instead of your equation (ii), I'll use one which is basically the same: $(1-x)^{-1}= 1 + x + x^2 + ...$ and I'm assuming that we can ignore all terms in $x$ of higher power than $x^5$.

So $\cos x = 1 - \frac{x^2}{2}+\frac{x^4}{24}= 1 - \Big(\frac{x^2}{2}-\frac{x^4}{24}\Big)$

$\Rightarrow (\cos x)^{-1} = 1 + \Big(\frac{x^2}{2}-\frac{x^4}{24}\Big) +\Big(\frac{x^2}{2}-\frac{x^4}{24}\Big)^2+...$

$= 1 +\frac{x^2}{2}-\frac{x^4}{24} + \Big(\frac{x^2}{2}\Big)^2+...$

$= 1 +\frac{x^2}{2}-\frac{x^4}{24} + \frac{x^4}{4}$

$= 1 + \frac{x^2}{2}+\frac{5x^4}{24}$

So $\tan x =(\sin x)(\cos x)^{-1}= \Big(x - \frac{x^3}{6}+\frac{x^5}{120}\Big) \Big(1 + \frac{x^2}{2}+\frac{5x^4}{24}\Big)$

$= x + x^3\Big(-\frac{1}{6}+\frac{1}{2}\Big)+ x^5\Big(\frac{5}{24}-\frac{1}{12}+\frac{1}{120}+...\Big)$

$= x +\tfrac13x^3 +\tfrac{2}{15}x^5$