The Student Room Group
Reply 1
Expansion?

I thought it was just sin/cos = tan, as a rule?
Reply 2
Think long division.
Reply 3
Try writing cos(x) * tan(x) = sin(x)

Now (1-x^2/2!+x^4/4!...) * (a+bx+cx^2+dx^3+ex^4+fx^5...) = (x-x^3/3!+x^5/5!)

You can equate coefficients e.g. the same number of x's have to appear on each side of the equation.

So, firstly there are no constants on the right, so a=0 (because a*1 in the expansion of the LHS would give you a constant).

Secondly, bx*1 = x, therefore b=1.

Thirdly, cx^2*1 + a*(-x^2/2!) = 0, therefore c=0 (because we found a=0).

Fourth, dx^3*1 + (bx)*(-x^2/2!) = -x^3/3!. Solving this gives you d = 1/3. (We found b=1).

And so on. Do it for as many terms as you feel necessary (perhaps up to x^5).


I must admit, I don't know a neater analytic way to do it.
Reply 4
i cant quite get to the right answer with that.

its meant to be tan x = 1 + x^3/3 + 2x^5/15

:confused:
Reply 5
Did you try long division?

Divide (x - x^3/3! + x^5/5! - ...) by (1 - x^2/2! + x^4/4! - ...).
Reply 6
i just tried the long division.

i managed to get the 2nd two terms right, but i keep getting x instead of 1 for the first term....
Reply 7
You should get an x.

tan(x) = x + x^3/3 + 2x^5/15 + ...
Reply 8
tribal_angel
i just tried the long division.

i managed to get the 2nd two terms right, but i keep getting x instead of 1 for the first term....

1 can't be the first term. tan 0 = 0
Reply 9
:eek: my textbook must be wrong, gah and to think the time i just spent on it
Original post by tribal_angel
:eek: my textbook must be wrong, gah and to think the time i just spent on it

omg, just spent and hour or two on this and had the same problem. my textbook is wrong and has that pesky 1 too :eek: