Problem Statement
Find a closed form for
sin
(
a
)
+
sin
(
a
+
b
)
+
sin
(
a
+
2
b
)
+
⋯
+
sin
(
a
+
(
n
−
1
)
b
)
{\displaystyle \sin(a)+\sin(a+b)+\sin(a+2b)+\cdots +\sin(a+(n-1)b)}
Note: A 'closed form' is not mathematically defined, but just means a simplified formula which does not involve '...', or a summation sign. In our problem, we should look for a formula that only involves variables
a
,
b
,
n
{\displaystyle a,b,n}
Method 1
To sum the series
sin
(
a
)
+
sin
(
a
+
b
)
+
sin
(
a
+
2
b
)
+
⋯
+
sin
(
a
+
(
n
−
1
)
b
)
=
S
{\displaystyle \sin(a)+\sin(a+b)+\sin(a+2b)+\cdots +\sin(a+(n-1)b)=S}
Multiply each term by
2
sin
(
b
2
)
{\displaystyle 2\sin \left({\tfrac {b}{2}}\right)}
Then we have
2
sin
(
a
)
sin
(
b
2
)
=
cos
(
a
−
b
2
)
−
cos
(
a
+
b
2
)
{\displaystyle 2\sin(a)\sin \left({\tfrac {b}{2}}\right)=\cos \left(a-{\tfrac {b}{2}}\right)-\cos \left(a+{\tfrac {b}{2}}\right)}
and similarly for all terms to
2
sin
(
a
+
(
n
−
1
)
b
)
sin
(
b
2
)
=
cos
(
a
+
(
2
n
−
3
)
b
2
)
−
cos
(
a
+
(
2
n
−
1
)
b
2
)
{\displaystyle 2\sin {\bigl (}a+(n-1)b{\bigr )}\sin \left({\tfrac {b}{2}}\right)=\cos \left(a+{\tfrac {(2n-3)b}{2}}\right)-\cos \left(a+{\tfrac {(2n-1)b}{2}}\right)}
Summing, we find that nearly all the terms cancel out and we are left with
2
S
sin
(
b
2
)
=
cos
(
a
−
b
2
)
−
cos
(
a
+
(
2
n
−
1
)
b
2
)
=
2
sin
(
a
+
(
n
−
1
)
b
2
)
sin
(
n
b
2
)
{\displaystyle 2S\sin \left({\tfrac {b}{2}}\right)=\cos \left(a-{\tfrac {b}{2}}\right)-\cos \left(a+{\tfrac {(2n-1)b}{2}}\right)=2\sin \left(a+{\tfrac {(n-1)b}{2}}\right)\sin \left({\tfrac {nb}{2}}\right)}
Hence
S
=
sin
(
a
+
(
n
−
1
)
b
2
)
sin
(
n
b
2
)
sin
(
b
2
)
{\displaystyle S=\sin \left(a+{\tfrac {(n-1)b}{2}}\right){\frac {\sin \left({\tfrac {nb}{2}}\right)}{\sin \left({\tfrac {b}{2}}\right)}}}
Similarly, if
C
=
cos
(
a
)
+
cos
(
a
+
b
)
+
cos
(
a
+
2
b
)
+
⋯
+
cos
(
a
+
(
n
−
1
)
b
)
{\displaystyle C=\cos(a)+\cos(a+b)+\cos(a+2b)+\cdots +\cos {\bigl (}a+(n-1)b{\bigr )}}
then
C
=
cos
(
a
+
(
n
−
1
)
b
2
)
⋅
sin
(
n
b
2
)
sin
(
b
2
)
{\displaystyle C=\cos \left(a+{\tfrac {(n-1)b}{2}}\right)\cdot {\frac {\sin \left({\tfrac {nb}{2}}\right)}{\sin \left({\tfrac {b}{2}}\right)}}}
Method 2
Consider the following sum
s
=
e
a
i
+
e
(
a
+
b
)
i
+
⋯
+
e
(
a
+
(
n
−
1
)
b
)
i
{\displaystyle s=e^{ai}+e^{(a+b)i}+\cdots +e^{(a+(n-1)b)i}}
Since
s
{\displaystyle s}
e
b
i
{\displaystyle e^{bi}}
s
=
e
a
i
(
e
n
b
i
−
1
)
e
b
i
−
1
=
e
a
i
e
n
b
i
2
(
e
n
b
i
2
−
e
−
n
b
i
2
)
e
b
i
2
(
e
b
i
2
−
e
−
b
i
2
)
{\displaystyle s={\frac {e^{ai}(e^{nbi}-1)}{e^{bi}-1}}={\frac {e^{ai}e^{\frac {nbi}{2}}(e^{\frac {nbi}{2}}-e^{-{\frac {nbi}{2}}})}{e^{\frac {bi}{2}}(e^{\frac {bi}{2}}-e^{-{\frac {bi}{2}}})}}}
s
=
sin
(
n
b
2
)
sin
(
b
2
)
⋅
e
(
a
+
(
n
−
1
)
b
2
)
i
{\displaystyle s={\frac {\sin \left({\tfrac {nb}{2}}\right)}{\sin \left({\tfrac {b}{2}}\right)}}\cdot e^{\left(a+{\tfrac {(n-1)b}{2}}\right)i}}
Therefore,
sin
(
a
)
+
sin
(
a
+
b
)
+
sin
(
a
+
2
b
)
+
⋯
+
sin
(
a
+
(
n
−
1
)
b
)
=
I
m
(
s
)
=
sin
(
a
+
(
n
−
1
)
b
2
)
⋅
sin
(
n
b
2
)
sin
(
b
2
)
{\displaystyle {\sin(a)+\sin(a+b)+\sin(a+2b)+\cdots +\sin {\bigl (}a+(n-1)b{\bigr )}={\rm {Im}}(s)=\sin \left(a+{\tfrac {(n-1)b}{2}}\right)\cdot {\frac {\sin \left({\tfrac {nb}{2}}\right)}{\sin \left({\tfrac {b}{2}}\right)}}}}
cos
(
a
)
+
cos
(
a
+
b
)
+
cos
(
a
+
2
b
)
+
⋯
+
cos
(
a
+
(
n
−
1
)
b
)
=
R
e
(
s
)
=
cos
(
a
+
(
n
−
1
)
b
2
)
⋅
sin
(
n
b
2
)
sin
(
b
2
)
{\displaystyle {\cos(a)+\cos(a+b)+\cos(a+2b)+\cdots +\cos {\bigl (}a+(n-1)b{\bigr )}={\rm {Re}}(s)=\cos \left(a+{\tfrac {(n-1)b}{2}}\right)\cdot {\frac {\sin \left({\tfrac {nb}{2}}\right)}{\sin \left({\tfrac {b}{2}}\right)}}}}
