# Get me going in the right direction on this and I'll give you a hug

schattenjaeger
cos(x) + cos3x + cos5x +………cos(2n -1)x=sin[2nx]/{2sinx}

Hint: Use Euler’s formula and the geometric progression formula.

Err, I know Euler's formula, or at least a version of it, but I don't see how that helps here, so yah, how's that work, since it obviously does?

## Answers and Replies

Staff Emeritus
Gold Member
You could start by replacing $cos[(2n-1)x]$ with $exp[i(2n-1)x]$, which will give you a geometric series.

schattenjaeger
ok, thanks! Now here's a good 'ol check my work question

y^3-x^2y=8, at the point (-3,1) is the slope 1 and the equation of the tangent line y=x+4?(that's what I got, I'm asking for confirmation)

EDIT: Nevermind, there was a typo in the problem on the teacher's site, 'cuz I found the same problem in the book with the point being (3,-1), which it has to be because (-3,1) doesn't solve that equation! Then the answer is y=x-4, which I checked with some cool graphy thing I downloaded. Three cheers!

and another one to get me started on

III.2: Find the largest box (with faces parallel to the coordinate axes) that can be inscribed within the ellipsoid:

(x^2)/4 + (y^2)/9 +(z^2)/25 =1

I'm assuming it's some type of lagrange multiplier problem, but I dunno...

Last edited:
schattenjaeger
Hah, yayyyy

For those wondering how to do the box problem, f(x,y,z)=8xyz, and g(x,y,z)=[thatequationfortheellipse], and then it's just good 'ol Lagrange multiplier stuff, with 4 unknowns and 4 equation, which happen to solve real nicely