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If $B$ be the exterior angle of a regular polygon of $n$ sides and $A$ is any constant, then prove that $\sin A + {\text{sin}}\left( {A + B} \right) + {\text{sin}}\left( {A + 2B} \right) + .........n$ terms$ = 0$

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Hint-This question can be solved by the formula when sin series is in Arithmetic Progression. In A.P two consecutive numbers in a series have common differences.

Now given that the regular polygon is $n - $ sided, also $A$is any constant and $B$ is an exterior angle and we have to prove that
$\sin A + {\text{sin}}\left( {A + B} \right) + {\text{sin}}\left( {A + 2B} \right) + .........n$ terms$ = 0$
Now we know that for a$n - $ sided polygon,
Sum of interior angle${\text{ = }}\left( {n - 2} \right)\pi $
Sum of exterior angle${\text{ = 2}}n\pi - \left( {n - 2} \right)\pi $
$
   = 2n\pi - n\pi + 2\pi \\
   = n\pi + 2\pi \\
  {\text{or }}B = \dfrac{{n\pi + 2\pi }}{n} \\
 $
For finding the value of $B$ we divide sum of exterior angle by$n$ because we want the value of exterior angle not the value of its sum
Now we have given, $\sin A + {\text{sin}}\left( {A + B} \right) + {\text{sin}}\left( {A + 2B} \right) + .........n$ terms
We can clearly see it is a A.P. with$a = A$ and$d = B = \dfrac{{n\pi + 2\pi }}{n}$
Now we know that the sum of sin series when angle is in A.P.${\text{ = }}\dfrac{{{\text{sin}}\left( {\dfrac{{nd}}{2}} \right)}}{{{\text{sin}}\left( {\dfrac{d}{2}} \right)}}{\text{sin}}\left( {\dfrac{{2a + (n - 1)d}}{2}} \right)$
Now putting the value of$a$ and$d$ we get,
Sum of sin series when angle is in A.P.
$
  {\text{ = }}\dfrac{{{\text{sin}}\left( {\dfrac{n}{2} \times \dfrac{{n\pi + 2\pi }}{n}} \right)}}{{{\text{sin}}\left( {\dfrac{{n\pi + 2\pi }}{{2n}}} \right)}} \times {\text{sin}}\left( {\dfrac{{2A + (n - 1) \times \dfrac{{n\pi + 2\pi }}{n}}}{2}} \right) \\
  {\text{ = }}\dfrac{{{\text{sin}}\left( {\dfrac{{n\pi + 2\pi }}{2}} \right)}}{{{\text{sin}}\left( {\dfrac{{n\pi + 2\pi }}{{2n}}} \right)}} \times {\text{sin}}\left( {\dfrac{{2A + (n - 1) \times \dfrac{{n\pi + 2\pi }}{n}}}{2}} \right) \\
 $
Now let us observe${\text{sin}}\left( {\dfrac{{n\pi + 2\pi }}{2}} \right)$
Or we can write it as${\text{sin}}\left( {\dfrac{{n + 2}}{2}} \right)\pi $
Now it is given that$n$ is the no. of sides of a regular polygon .Therefore it is an integer.
$
  {\text{or }}n \in I \\
  {\text{or }}n + 2 \in I \\
  {\text{or }}\dfrac{{n + 2}}{2} \in I \\
 $
Or we can say
${\text{sin}}\left( {\dfrac{{n + 2}}{2}} \right)\pi {\text{ = sin}}\left( {m\pi } \right){\text{ }}m \in I$
And we know that${\text{sin}}\left( {m\pi } \right) = 0{\text{ }}m \in I$
$
   \Rightarrow {\text{ = }}\dfrac{{{\text{sin}}\left( {\dfrac{{n\pi + 2\pi }}{2}} \right)}}{{{\text{sin}}\left( {\dfrac{{n\pi + 2\pi }}{{2n}}} \right)}} \times {\text{sin}}\left( {\dfrac{{2A + (n - 1) \times \dfrac{{n\pi + 2\pi }}{n}}}{2}} \right) \\
   = \dfrac{{\sin \left( {m\pi } \right)}}{{\sin \left( {\dfrac{{m\pi }}{n}} \right)}} \times \sin \left( {2A + \dfrac{{n - 1}}{n}m\pi } \right) \\
 $
Now we know that${\text{sin}}\left( {m\pi } \right) = 0$
$
   \Rightarrow {\text{ }}\sin A + {\text{sin}}\left( {A + B} \right) + {\text{sin}}\left( {A + 2B} \right) + .........n{\text{ terms}} = \dfrac{{\sin \left( {m\pi } \right)}}{{\sin \left( {\dfrac{{m\pi }}{n}} \right)}} \times \sin \left( {2A + \dfrac{{n - 1}}{n}m\pi } \right) \\
   = 0 \\
 $
Hence Proved

Note: Whenever we face such types of problems the key concept is that we should know the formula when the sin series is in A.P. Like in this question it is given that the polygon is regular and we write the formula for sum of interior as well as exterior angle then we see that it is in A.P. and we know the formula when Sin series is in A.P. and thus we prove it.
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