Trigonometry+ 03: UNIZOR.COM - Math+ & Problems - Trigonometry

Notes to a video lecture on http://www.unizor.com

Trigonometry+ 03


Problem A

Find the sums
S_sin(x,y,n)= Σ_k∈[0,n]sin(x+k·y)
S_cos(x,y,n)= Σ_k∈[0,n]cos(x+k·y)
as algebraic expression with arguments x, y and n.


Solution A1

Let's start with a sum of sines.
Recall the formula for a cosine of a sum of two angles
cos(a+b) =
= cos(a)·cos(b) − sin(a)·sin(b)
From this formula it's easy to derive another trigonometric identity formula
2·sin(a)·sin(b) =
= cos(a−b)−cos(a+b)

Let's use this formula to transform our series into only a couple of members.
Multiply S_sin (that is, each member of this series) by 2·sin(y/2) and use the trigonometric identity above.

Then after this multiplication by 2·sin(y/2) the k^th member of S_sin would look like
2·sin(x+k·y)·sin(y/2) =
= cos(x+k·y−y/2) −
− cos(x+k·y+y/2) =
= cos(x+(2k−1)·y/2) −
− cos(x+(2k+1)·y/2)

Thus, each member of our series is converted into a pair of cosines with a minus in-between:
2·S_sin(x,y,n)·sin(y/2) =
= Σ_k∈[0,n][cos(x+(2k−1)·y/2) −
− cos(x+(2k+1)·y/2)] =
= [cos(x−y/2)−cos(x+y/2)] +
+ [cos(x+y/2)−cos(x+3y/2)] +
+ [cos(x+3y/2)−cos(x+5y/2)] +
etc. and the last two members of a series are
+ [cos(x+(2n−3)/2) −
− cos(x+(2n−1)y/2)] +
+ [cos(x+(2n−1)/2) −
− cos(x+(2n+1)y/2)]

It's easy to see that in this series every members, except the first and the last, cancel each other and the result is
2·S_sin(x,y,n)·sin(y/2) =
= cos(x−y/2) −
− cos(x+(2n+1)·y/2)

Therefore,
S_sin(x,y,n) =
= [cos(x−y/2) −
− cos(x+(2n+1)·y/2)] /
/ [2·sin(y/2)]

The sum of cosines can be treated analogously.
Recall the formula for a sine of a sum of two angles
sin(a+b) =
= sin(a)·cos(b) + cos(a)·sin(b)
From this formula it's easy to derive another trigonometric identity formula
2·cos(a)·sin(b) =
= sin(a+b)−sin(a−b)

Let's use this formula to transform our series into only a couple of members.
First, we multiply all members of S_cos (that is, each member of this series) by 2·sin(y/2). and use the trigonometric identity above.
2·S_cos(x,y,n)·sin(y/2) =
= Σ_k∈[0,n][sin(x+(2k+1)·y/2) −
− sin(x+(2k−1)·y/2)] =
= sin(x+(2n+1)·y/2) −
− sin(x−y/2)

Therefore,
S_cos(x,y,n) =
= [sin(x+(2n+1)·y/2) −
− sin(x−y/2)] /
/ [2·sin(y/2)]

Of cause, both formulas for S_sin and S_cos are correct only if sin(y/2)≠0, that is y≠2πN, where N - any integer number.
If, however, y=2πN, then sin(x+k·y)=sin(x) and cos(x+k·y)=cos(x), so
S_sin = (n+1)·sin(x)
S_cos = (n+1)·cos(x)


Solution A2

A different and, arguably, more elegant approach to solve this problem would be use the Euler formula
e^i·φ = cos(φ)+i·sin(φ)
[See lecture on UNIZOR.COM - Math 4 Teens - Trigonometry - Complex Numbers and Trigonometry - Euler's Formula]

Using it, we can express cos(x+k·y) and sin(x+k·y) as real and imaginary parts of
e^i·(x+k·y) = e^i·x·(e^i·y)^k

The sum of these expression by k from 0 to n can be calculated as a geometric series with the first term a=e^i·x and factor r=e^i·y.
[See lecture on UNIZOR.COM - Math 4 Teens - Algebra - Sequence and Series - Progression Series]
This sum in terms of a and r is
S = a·(rⁿ⁺¹−1)/(r−1)

Therefore, our sum is equal to
Σ_k∈[0,n]e^i·x·(e^i·y)^k =
= e^i·x·[(e^i·y)ⁿ⁺¹−1] /(e^i·y−1) =
= [e^{i·(x+y·(n+1))}−e^i·x] /(e^i·y−1)

What remains is to convert this expression into canonical representation of complex numbers as A+B·i. Then A will represent the sum of cosines, and B will represent a sum of sines.

First, get rid of i in the denominator by multiplying both numerator and denominator by (e^−i·y−1).
In the denominator the result will be
(e^i·y−1)·(e^−i·y−1) =
= e⁰−e^−i·y−e^i·y+1 =
= 2 − e^i·y − e^−i·y =
= 2 − cos(y) − i·sin(y) −
− cos(−y) − i·sin(−y) =
= 2 −2cos(y) = 4·sin²(y/2)
As we see, there is only real part in the denominator, an imaginary terms cancel each other.

In the numerator we will have
[e^{i·(x+y·(n+1))}−e^i·x]·(e^−i·y−1) =
= e^i·(x+y·n) − e^{i·(x+y·(n+1))} −
− e^i·(x−y) + e^i·x =
= cos(x+y·n)−cos(x+y·(n+1))−
−cos(x−y) + cos(x) +
+ i·[sin(x+y·n)−sin(x+y(n+1))−
−sin(x−y)+sin(x)]

Let's deal with real part that gives the sum of cosines.
We will use another trigonometric identity which can be easily proved using
α=½(α+β)+½(α−β) and
β=½(α+β)−½(α−β):
cos(α) − cos(β) =
= −2sin(½(α+β))·sin(½(α−β))
Using this identity, the numerator for a sum of cosines is
cos(x+y·n)−cos(x+y·(n+1))−
−cos(x−y) + cos(x) =
= 2·sin(x+½(2n+1)·y)·sin(½y) −
− 2·sin(x−½y)·sin(½y) =
= 2·sin(½y)·
·[sin(x+½(2n+1)·y) −
− sin(x−½y)]

Combining this numerator and calculated above denominator 4·sin²(y/2), we obtain
S_cos(x,y,n) =
= [sin(x+(2n+1)·y/2) −
− sin(x−y/2)]·
·2·sin(y/2) / [4·sin²(y/2)] =
= [sin(x+(2n+1)·y/2) −
− sin(x−y/2)] /
/ [2·sin(y/2)]
which corresponds completely to Solution B1.

Similarly, the imaginary part of a numerator can be used to calculate the sum of sines.
We will use yet another trigonometric identity, which can be derived analogously to a difference of cosines above:
sin(α) − sin(β) =
= 2sin(½(α−β))·cos(½(α+β))
Using this identity, the numerator for a sum of sines is
sin(x+y·n)−sin(x+y(n+1))−
−sin(x−y)+sin(x) =
=−2·sin(½y)·cos(x+½(2n+1)·y)+
+ 2·sin(½y)·cos(x−½y) =
= 2·sin(½y)·
·[cos(x−½y) −
− cos(x+½(2n+1)·y)]

Combining this numerator and calculated above denominator 4·sin²(y/2), we obtain
S_sin(x,y,n) =
= [cos(x−y/2) −
− cos(x+(2n+1)·y/2)]·
·2·sin(y/2) / [4·sin²(y/2)] =
= [cos(x−y/2) −
− cos(x+(2n+1)·y/2)] /
/ [2·sin(y/2)]
which corresponds completely to Solution B1.