# S.L. Loney Some Simple Trigonometrical Series Solutions (Chapter 19)

SL Loney Plane Trigonometry Solutions for Chapter 19 ‘Some Simple Trigonometrical Series’ are created by our subject matter experts as per the latest IIT JEE syllabus. These solutions are an essential guide if you want to establish a grip over the Trigonometrical Series for competitive exams like JEE. Our SL Loney Solutions for the chapter will help you understand how by using arithmetic progression for angles of a triangle we can find the sum of the series of an angle. After going through our SL Loney solutions, you will also understand concepts such as the sum of the sine series of angles, the sum of the cosine series of angles, the sum of cube of the sine/ cosine series of angles.

SL Loney Plane Trigonometry Some Trigonometrical Series chapter inter-relates functions with trigonometry and can be immensely helpful in preparing for competitive exams like JEE Main and JEE Advanced. Trigonometrical Series also has a useful significance in measuring the vibrations or propagation of heat that you would study in higher education. There are a total of 27 questions covered in one exercise in this chapter. These questions would help you learn to derive the formula for the sum of the sine/ cosine series of angles to solve problems.

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## Important Topics for SL Loney Plane Trigonometry Solutions Chapter 19: Some Simple Trigonometrical Series

**Sum of the Sine of a Series of Angles: **Considering angle being in arithmetic progression, let us derive an equation for the sum of the sine of a series of angles.

Let angles be a, (a + b), (a + 2b), (a + 3B)… (a + (n – 1) b)

The sum of series is

S = sin (a) + sin (a + b) + sin (a + 2b) + … + sin (a + (n -1) b)

Multiplying each term by

2 sin (b/2)

We get

2 sin (a) sin (b/2) = cos (a – (b/2) – cos [a + (b/2)]

2 sin (a + (n – 1) b) sin (b/2) = cos [a – ((2n – 3) b/ 2)] – cos [a + ((2n – 3) b/ 2)]

After summing all of them we find that nearly all terms cancel out and we are left with:

2 S sin (b/2) = cos [a – (b/2)] – cos [a + ((2n – 1) b/2)] = 2 sin [a + ((n – 1) b / 2)] sin (nb/2)

Hence we can say:

**Sum of the cosine of a series of angles: **Considering angle being in arithmetic progression, let us derive an equation for the sum of the cosine of a series of angles.

Let angles be a, (a + b), (a + 2b), (a + 3B)… (a + (n – 1) b)

The sum of series is

C = cos (a) + cos (a + b) + cos (a + 2b) + … + cos (a + (n -1) b)

Multiplying each term by

2 sin (b/2)

We get

2 cos (a) sin (b/2) = sin (a + (b/2) – sin [a – (b/2)]

2 cos (a + (n – 1) b) sin (b/2) = sin [a + ((2n – 3) b/ 2)] – sin [a – ((2n – 3) b/ 2)]

After summing all of them we find that nearly all terms cancel out and we are left with:

2 C sin (b/2) = 2 cos [a + ((n – 1) b / 2)] sin (nb/2)

Hence we can say:

**Sum of the sine/ cosine of a series of angles becomes zero: **When (nb/2) is equal to any multiple of π, then the sum of the sine/ cosine series of an angle becomes 0. Let us see how:

When (nb/2) = p π, where p is any integer,

That is b = π * (2 π/n)

Hence the sum of the sine/ cosine series of angles vanishes as the common difference of angle is any multiple of (2 π/n).

### Exercise Discussion for SL Loney Plane Trigonometry Solutions Chapter 19: Some Simple Trigonometrical Series

- Some Simple Trigonometrical Series of SL Loney Plane Trigonometry solutions includes 1 exercise with 27 problems.
- The exercise questions are based on topics of finding the angular points of the polygon, the sum of the sine series, the sum of the cosine series, the sum of cubes of the sine/ cosine series, and to prove some equations.
- This chapter also includes problems based on the most important topic of the trigonometrical series, that is the sum of the sine/ cosine series.

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