# Summation of a trigonometric function

• Supernova123
In summary, by considering ∑z2n-1, where z=eiθ, it can be shown that Σcos(2n-1)θ=sin(2Nθ)/2sinθ.
Supernova123

## Homework Statement

By considering ∑z2n-1, where z=e, show that Σcos(2n-1)θ=sin(2Nθ)/2sinθ. (Σ means summation from 1 to N)

## Homework Equations

Just a guess. S=a(1-r^n)/(1-r)

## The Attempt at a Solution

I was thinking this but it doesn't seem to work very well. Σz2n-1=Σ(cosθ+isinθ)2n-1=Σ(cos(2n-1)θ+isin(2n-1)θ). Rearranging,
Σcos(2n-1)θ=Σ(cosθ+isinθ)2n-1-Σsin(2n-1)θ
=(cosθ+isinθ)(1-(cosθ+isinθ)2N/(1-(cosθ+isinθ)2) - isinθ(1-(isinθ)2N/(1-(isinθ)^2)

I suggest not rewriting in terms of sines and cosines until after doing the summation.

Okay, so I got:
Σcos(2n-1)θ = ∑z2n-1 - Σsin(2n-1)θ
=z(1-z2N)/(1-z2) - isinθ(1-(isinθ)2)/(1-(isinθ)2)

=(z-z2N+1)/(1-z2) - (isinθ - isin(2N+1)θ)/(1 - isin2θ)

=(z - izsin2θ - z2N+1 + iz2N+1sin2θ - isinθ + isin(2N+1)θ + iz2sinθ - iz2sin(2N+1)θ)/(1 - z2 - isin2θ + iz2sin2θ)

=(cosθ + isinθ - icosθsin2θ + sinθsin2θ -cos(2N+1)θ -isin(2N+1)θ + icos(2N+1)θsin2θ - sin(2N+1)sin2θ - isinθ + isin(2N+1)θ + icos2θsinθ - sin2θsinθ - icos2θsin(2N+1)θ + sin2θsin(2N+1)θ)/(1 - cos2θ - isin2θ - isin2θ + icos2θsin2θ - sin2θsin2θ)

=(cosθ - icosθsin2θ - cos(2N+1)θ + icos(2N+1)θsin2θ)/(1 - cos2θ - 2isin2θ + icos2θsin2θ - sin2θsin2θ) and I'm stuck :(

Supernova123 said:

## Homework Statement

By considering ∑z2n-1, where z=e, show that Σcos(2n-1)θ=sin(2Nθ)/2sinθ. (Σ means summation from 1 to N)

## Homework Equations

Just a guess. S=a(1-r^n)/(1-r)

## The Attempt at a Solution

I was thinking this but it doesn't seem to work very well. Σz2n-1=Σ(cosθ+isinθ)2n-1=Σ(cos(2n-1)θ+isin(2n-1)θ). Rearranging,
Σcos(2n-1)θ=Σ(cosθ+isinθ)2n-1-Σsin(2n-1)θ
=(cosθ+isinθ)(1-(cosθ+isinθ)2N/(1-(cosθ+isinθ)2) - isinθ(1-(isinθ)2N/(1-(isinθ)^2)

If ##z = e^{i \theta}## then ##z^{2n-1} = e^{(2n-1) i \theta} = \cos((2n-1) \theta) + i \sin((2n-1) \theta)##.

I found my mistake. I wrongly assumed that (sinθ)^n=sin(nθ). Here's my solution:
Σcos(2n-1)θ=Re(Σz2n-1)
=Re(e(1-e2Niθ)/(1-e2iθ))
=Re((1-e2Niθ)/(e-iθ-e))
=Re((1-cos(2Nθ)-isin(2Nθ)/(-2isinθ))
=Re((i-icos(2Nθ)+sin(2Nθ)/(2sinθ))
=sin(2Nθ)/2sinθ

## 1. What is the summation of a trigonometric function?

The summation of a trigonometric function refers to the process of adding up all the values of the function for a given range of input values. It is often used in calculus and other mathematical applications to find the total value of a function over a specific interval.

## 2. How is the summation of a trigonometric function calculated?

The summation of a trigonometric function is calculated by first evaluating the function at each of the input values in the given range. These values are then added together to find the total sum. In some cases, this can be done analytically using mathematical formulas, while in other cases, numerical methods or computer programs may be used.

## 3. What are some common trigonometric functions used in summation?

Some of the most commonly used trigonometric functions in summation include sine, cosine, tangent, and their inverse functions. These functions are often used in combination with other mathematical functions to model real-world phenomena and solve complex problems in fields like physics, engineering, and economics.

## 4. Can the summation of a trigonometric function be done for any interval?

Yes, the summation of a trigonometric function can be done for any interval as long as the function is defined and continuous over that interval. However, the complexity of the summation may vary depending on the properties of the function and the chosen interval.

## 5. How is the summation of a trigonometric function related to the area under the curve?

The summation of a trigonometric function is closely related to the area under the curve of that function. In fact, the summation can be interpreted as an approximation of the area under the curve by breaking it into smaller trapezoidal or rectangular shapes. As the size of these shapes decreases, the summation becomes a more accurate representation of the area under the curve.

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