# Sum of Finite Sine Series

In trigonometry, it is easy to find the sum of finite series of sine of angles using our heroic calculator. What about if we are ask to find an expression to represent such series? To answer that question let me share a formula to solve it without using calculus.

Formula:

$\sum\limits_{i=1}^n sin(ix)=\displaystyle\frac{sin(\frac{nx}{2})sin(\frac{n+1}{2})x}{sin(\frac{x}{2})}$

Let’s work out some examples and we will use this formula.

Sample Problem 1:

Express the sum of sin1° + sin2° + sin3° + . . . + sin180° as a single expression.

Solution:

We can rewrite the finite sum as $\sum\limits_{i=1}^{180} sin(i)$

Using our formula above we have,

$\sum\limits_{i=1}^n sin(ix)=\displaystyle\frac{sin(\frac{nx}{2})sin(\frac{n+1}{2})x}{sin(\frac{x}{2})}$

where n=180, x=1

$\sum\limits_{i=1}^{180} sin(i)=\displaystyle\frac{sin(\frac{(180)(1)}{2})sin(\frac{180+1}{2})1}{sin(\frac{1}{2})}$

$\sum\limits_{i=1}^{180} sin(i)=\displaystyle\frac{sin(90)sin(\frac{181}{2})}{sin(\frac{1}{2})}$

Sin90°=1

$\sum\limits_{i=1}^{180} sin(i)=\displaystyle\frac{ sin(\frac{181}{2})}{sin(\frac{1}{2})}$

To reduce this further, sin(181/2)=sin90.5

Applying sum of angle of sine we have,

sin(90+0.5)=sin90cos0.5+cos90sin0.5

sin(90+0.5) = cos0.5

$\sum\limits_{i=1}^{180} sin(i)=\displaystyle\frac{ cos(\frac{1}{2})}{sin(\frac{1}{2})}$

$\sum\limits_{i=1}^{180} sin(i)=cot\displaystyle\frac{1}{2}$

sin1° + sin2° + sin3° + . . . + sin180°=cot (1/2)°

Sample Problem 2:

Express the sum of sin2° + sin4° + sin6° + . . . + sin180° as a single expression.

Solution:

We can rewrite the expression

sin2° + sin4° + sin6° + . . . + sin180°=  $\sum\limits_{i=1}^{90} sin(2i)$

$\sum\limits_{i=1}^{90}sin(2i)=\displaystyle\frac{sin(\frac{nx}{2})sin(\frac{n+1}{2})x}{sin(\frac{x}{2})}$

Where n=90 , x=2

$\sum\limits_{i=1}^{90}sin(2i)=\displaystyle\frac{sin(\frac{90(2)}{2})sin(\frac{90+1}{2})2}{sin(\frac{2}{2})}$

$\sum\limits_{i=1}^{90}sin(2i)=\displaystyle\frac{sin(90)sin(91)}{sin(1)}$

but sin90=1.

$\sum\limits_{i=1}^{90}sin(2i)=\displaystyle\frac{sin(91)}{sin(1)}$

Also, sin91 = sin(90+1) = sin90cos1+cos90sin1 = cos1

$\sum\limits_{i=1}^{90}sin(2i)=\displaystyle\frac{cos(1)}{sin(1)}$

$\sum\limits_{i=1}^{90}sin(2i)=cot1$

sin2° + sin4° + sin6° + . . . + sin180°= cot1°

### Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.

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