Sum of Sine or Cosine of N Angles in A.P.

In mathematics, arithmetic expressions can form a series by adding a constant value in every term. This series is called an arithmetic progression series. We can find the appearance of trigonometric functions and the sum of trigonometric functions when the angles are given in the sequence of arithmetic progression. In that case, the sum of sine values or cosine values of N angles can be considered as the sum of sine or cosine of N angles in A.P. As mentioned below, we can find an argument for relations of the sum provided by Judy Holdener:

k = 1Ncos(2πk /N) = 0 and k = 1Ncos(2πk /N) = 0  

These arguments come in the realm of complex numbers. We can also say that this relation is formed because of formulas. These formulas don’t need to be proven using complex numbers. We can also give these arguments as a theorem. 

Let’s say, a, d ∈ R where d is not equal to zero and n is a positive integer number, that the formula of these relations can be written as follows.

k = 1n-1sin(a + kd) = [sin(nd/2)/sin(d/2)]cos(a + (n-1)d/2)

And

 k = 1n-1cos(a + kd) = [sin(nd/2)/sin(d/2)]cos(a + (n-1)d/2)

To prove this theorem, we will start by rearranging the sum. Terms in the current iteration can cable the term in the last iteration, and this also explains how to get rid of the sum of trigonometric functions. We can also call it a telescoping series.

Let’s start with the cosine sum.

Sum of Cosine of N angels in A.P.

Let’s say sin(d/2) = 0. If this is right, we can say that d can be a multiple of 2ℼ. And the expression of cosine arithmetic progression is:

cos(a) = cos(a+ d) = cos(a+2d)…..

And

n=0N-1 cos( a + nd) = Ncosa 

By the angle sum rules we know that,

sin(a+b) = sinasinb + cosasinb

sin(a-b) = sinacosb-cosasinb

And 

sin(a+ b) – sin(a-b) = 2cos a sin b

Now, 

 C= n=0N-1cos(a +nd) ………(1)

This coverage is based on the condition where sin(d/2) = 0. Now, if sin(d/2) 0, then we can multiply 2sin(d/2) on both sides of equation one as follows:

2sin(d/2)C =n=0N-12cos(a +nd)sin(d/2) 

Now, we can use the angle sum rule. Let a = (a+nd) and b = 1/2d

2sin(d/2) C =n=0N-1[sin{ a +(n+1/2)/d} -sin{ a +(n-1/2)/d} 

We can expand this sum as follows:

2sin(d/2) C =[sin(a+d/2)-sin(a-d/2)]+[sin(a+3d/2)-sin(a-d/2)]+

….. [sin(a+(N-3/2)d)-sin(a +(N-5d/2)]+

[sin(a+(N-1/2)d)-sin(a +(N-3d/2)]

We can also call this series a telescopic series. The reason behind calling it telescopic is that the second term of the expression cancels or subtracts the first term of the previous iteration in each iteration. Only the first term from the last iteration and the last term from the first iteration are left in the expression.

The outcome of expansion is as follows:

2sin(d/2) C= sin(a+(N-1/2)d) -sin(a-d/2) 

Now, if we go with the angle sum rule.

sin(a+b)-sin(a-b)=2cosa cosb

Then we will get the final expression as:

2sin(d/2) C= 2cos(a+(N-1)d/2)sin(Nd/2)

Here we solve C to finish the proof. Now let’s move on towards the sine sum. Since it is identical, we are performing it after cosine sum.

Sum of Sine of N angels in A.P

Here we also know the angle sum rule.

cos(a+b)=cosacosb-sinasinb

cos(a-b)=cosacosb+sinasinb

And,

 cos(a+b)-cos(a-b)= -2sinasinb

Now the equation of sine sum in A.P is 

S = n=oN-1sin(a+nd)

If sin(d/2) 0, we can multiply -2sin(d/2) on both sides.

-2sin(d/2)S = n=oN-1-2sin(a+nd)sin(d/2) 

Let’s use the above-given angle sum rule.

-2sin(d/2)S=n=oN-1[cos(a+(n+1/2)d)-cos(a+(n-1/2)d)] .

= cos[a+(n+1/2)d)-cos(a+(n-1/2)d)]

= -2sin(a+(n-1)d/2)sin(Nd/2)

This is the solution for the sine of N values in arithmetic progression.

Conclusion 

We have introduced the sum of sine or cosine of N angles in arithmetic progression. We can perform the sum of sine/cosine of N angles in A.P. We also discussed some examples to explain how to do the sum of the cosine of N angle and the sum of the sin of N angles in arithmetic progression.