Lesson Calculating the sum 1*sin(1°) + 2*sin(2°) + 3*sin(3°) + . . . + 180*sin(180°)

Source code of 'Calculating the sum 1sin(1°) + 2sin(2°) + 3sin(3°) + . . . + 180sin(180°)'

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<H2>Calculating the sum 1*sin(1°) + 2*sin(2°) + 3*sin(3°) + . . . + 180*sin(180°)</H2>

The problem below is above the average school Math level.
It is a typical Math Circle level problem.

<H3>Problem 1</H3>Find the sum of 1*sin(1°) + 2*sin(2°) + 3*sin(3°) + . . . + 180*sin(180°).

<B>Solution</B>

<pre>
Let  R = 1*sin(1°) + 2*sin(2°) + 3*sin(3°) + . . . + 180*sin(180°).

It is even more convenient for me to write

R = 0*sin(0°) + 1*sin(1°) + 2*sin(2°) + 3*sin(3°) + . . . + 180*sin(180°)  by adding  0*sin(0°) = 0  as the first term.

Then I can write

R =   0*sin(0°)   + 1*sin(1°)     + 2*sin(2°)    + 3*sin(3°)     + . . . + 180*sin(180°)     (1)

R = 180*sin(180°) + 179*sin(179°) + 178*sin(178°) + 177*sin(177°) + . . . + 0*sin(0°).        (2)

Note that sin(0°) = sin(180°);  sin(1°) = sin(179°);  sin(2°) = sin(178°);  sin(3°) = sin(177°);  and so on.   
Therefore, adding  (1)  and  (2)

2R = 180*(sin(0°) + sin(1°) + sin(2°) + sin(3°) + . . . + sin(90°)).

So, our sum R <U>is the same as</U>

R = 90*(sin(0°) + sin(1°) + sin(2°) + sin(3°) + . . . + sin(90°)).

Now, let z = cos(1°) + i*sin(1°)  is this complex number.

Then, according to the de Moivre's formula

{{{z^2}}} = cos(2°) + i*sin(2°)

{{{z^3}}} = cos(3°) + i*sin(3°),

.  .  .  .  .  .

{{{z^90}}} = cos(90° + i*sin(90°),

and the sum R  is 90 times the <U>imaginary part</U>  of the sum of the geometric progression

Q = 1 + {{{z}}} + {{{z^2}}} + {{{z^3}}} + {{{z^4}}} + . . . + {{{z^90}}}.

The sum of the geometric progression is

{{{(z^91-1)/(z-1)}}} = {{{(cos(91^o) + i*sin(91^o) - 1)/(cos(1^o) + i*sin(1^o) - 1)}}} = now I will work to extract the imaginary part = {{{((cos(91^o) -1) + i*sin(91^o))/((cos(1^o)-1) + i*sin(1^o))}}} =

= {{{((cos(91^o) -1) + i*sin(91^o))/((cos(1^o)-1) + i*sin(1^o))}}} . {{{((cos(1^o)-1) -  i*sin(1^o))/((cos(1^o)-1) - i*sin(1^o))}}}

So, the imaginary part has the DENOMINATOR  (cos(1°)-1)^2+sin^2(1°) = 2-2*cos(1°) = 2*(1-cos(1°)).

The imaginary part has the NUMERATOR  sin(91°)*(cos(1°)-1) - (cos(91°)-1)*sin(1°) = sin(91°)*cos(1°) - cos(91°)*sin(1°) - sin(91°) + sin(1°) =

= sin(90°) - sin(91°) + sin(1°) = 1 - cos(1°) + sin(1°).

Therefore, the answer is:

the sum  R = 1*sin(1°) + 2*sin(2°) + 3*sin(3°) + . . . + 180*sin(180°) = {{{90*((1 - cos(1^o) + sin(1^o))/(2*(1-cos(1^o))))}}}.

You can transform it further as you want and/or as you need.

You can even get the numerical value, using your calculator.
</pre>

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Lesson Calculating the sum 1*sin(1°) + 2*sin(2°) + 3*sin(3°) + . . . + 180*sin(180°)

Source code of 'Calculating the sum 1*sin(1°) + 2*sin(2°) + 3*sin(3°) + . . . + 180*sin(180°)'

Lesson Calculating the sum 1sin(1°) + 2sin(2°) + 3sin(3°) + . . . + 180sin(180°)

Source code of 'Calculating the sum 1sin(1°) + 2sin(2°) + 3sin(3°) + . . . + 180sin(180°)'