Simplify:

Question

(cosx + cos2x + cos3x + cos4x + cos5x + cos6x) / (sinx + sin2x + sin3x + sin4x + sin5x + sin6x)

into a single Cotangent function.

Using the sum-to-products, I was able to get remove some of the addition in attempts to get full multiplication, but I could not get the 2 out of the denominator for the 5x or 6x.

Am I going about this the wrong way? Any ideas?

Answer 1

Try multiplying AND dividing each series by 2sin(x/2), expand and apply the sum/product formulas.
Many of the terms should cancel and leave you something simple to work with.

Answer 2

Here's the whole deal:

Let f(x) = (cos(x)+cos(2x)+...cos(6x)) / (sin(x)+...+sin(6x))

Multiply the denominator by 2sin(x/2) to get six products of 2sin(x/2)*sin(nx) where n=1 to 6.
Using the sum and product formula, you should get:
denominator =
cos(x/2)-cos(3x/2)+
cos(3x/2)-cos(5x/2)+
cos(5x/2)-cos(7x/2)+
cos(7x/2)-cos(9x/2)+
cos(9x/2)-cos(11x/2)+
cos(11x/2)-cos(13x/2)
= cos(x/2)-cos(13x/2)
= 2sin(7x/2)sin(3x)

Similarly, multiply the numerator by 2sin(x/2) gives
numerator = 2cos(7x/2)sin(3x)

On cancelling sin(3x), we get the simplified expression as:
cot(7x/2)

Check: for x=π/4,
cot(7x/2)=cot(7π/8)=-2.41421...

cos(π/4) + cos(2π/4) + cos(3π/4) + cos(4π/4) + cos(5π/4) + cos(6π/4)
= r+0-r-1-r+0
= -(1+sqrt(2)/2)

sin(π/4) + sin(2π/4) + sin(3π/4) + sin(4π/4) + sin(5π/4) + sin(6π/4)
= r+1+r+0-r-1
= r
= sqrt(2)/2

f(π/4)
= -(1+sqrt(2)/2) / sqrt(2)/2
= -2.41421...