Asked by Jeff
Simplify:
(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?
(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?
Answers
Answered by
MathMate
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.
Many of the terms should cancel and leave you something simple to work with.
Answered by
MathMate
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...
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...
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