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From the identity
sin
(
∑
i
=
1
∞
θ
i
)
=
∑
odd
k
≥
1
(
−
1
)
(
k
−
1
)
/
2
∑
A
⊆
{
1
,
2
,
3
,
…
}
|
A
|
=
k
(
∏
i
∈
A
sin
θ
i
∏
i
∉
A
cos
θ
i
)
{\displaystyle \sin \left(\sum _{i=1}^{\infty }\theta _{i}\right)=\sum _{{\text{odd }}k\geq 1}(-1)^{(k-1)/2}\sum _{\begin{smallmatrix}A\subseteq \{\,1,2,3,\dots \,\}\\\left|A\right|=k\end{smallmatrix}}\left(\prod _{i\in A}\sin \theta _{i}\prod _{i\not \in A}\cos \theta _{i}\right)}
one can infer that
sin
(
n
θ
)
=
∑
odd
k
∈
{
1
,
…
,
n
}
(
−
1
)
(
k
−
1
)
/
2
(
n
k
)
sin
k
θ
cos
n
−
k
θ
.
{\displaystyle \sin(n\theta )=\sum _{{\text{odd }}k\in \{1,\dots ,n\}}(-1)^{(k-1)/2}{n \choose k}\sin ^{k}\theta \cos ^{n-k}\theta .}
Letting x = nθ , we have
sin
x
=
∑
odd
k
∈
{
1
,
…
,
n
}
(
−
1
)
(
k
−
1
)
/
2
(
n
k
)
sin
k
θ
cos
n
−
k
θ
=
∑
odd
k
∈
{
1
,
…
,
n
}
[
(
−
1
)
(
k
−
1
)
/
2
(
n
(
n
−
1
)
⋯
(
n
−
k
+
1
)
k
!
)
(
x
n
)
k
⋅
(
sin
(
x
/
n
)
x
/
n
)
k
cos
n
−
k
(
x
/
n
)
]
=
∑
odd
k
∈
{
1
,
…
,
n
}
[
(
−
1
)
(
k
−
1
)
/
2
x
k
k
!
⋅
n
(
n
−
1
)
⋯
(
n
−
k
+
1
)
n
k
⋅
(
sin
(
x
/
n
)
x
/
n
)
k
cos
n
−
k
(
x
/
n
)
]
{\displaystyle {\begin{aligned}\sin x&=\sum _{{\text{odd }}k\in \{1,\dots ,n\}}(-1)^{(k-1)/2}{n \choose k}\sin ^{k}\theta \cos ^{n-k}\theta \\[12pt]&=\sum _{{\text{odd }}k\in \{1,\dots ,n\}}\left[(-1)^{(k-1)/2}\left({n(n-1)\cdots (n-k+1) \over k!}\right)\left({x \over n}\right)^{k}\right.\\&{}\qquad \qquad \qquad \qquad \qquad \qquad \qquad \left.{}\cdot \left({\sin(x/n) \over x/n}\right)^{k}\cos ^{n-k}(x/n)\right]\\[12pt]&=\sum _{{\text{odd }}k\in \{1,\dots ,n\}}\left[(-1)^{(k-1)/2}\ {x^{k} \over k!}\cdot {}\right.\\&{}\qquad \qquad \left.{n(n-1)\cdots (n-k+1) \over n^{k}}\cdot \left({\sin(x/n) \over x/n}\right)^{k}\cos ^{n-k}(x/n)\right]\end{aligned}}}
Having reached this point, Euler said that if n is an infinitely large integer, then the three factors on the last line above are equal to 1. In modern language, we would say that as n → ∞, the factors on the last line approach 1 and we get
sin
x
=
∑
odd
k
≥
1
(
−
1
)
(
k
−
1
)
/
2
x
k
k
!
.
{\displaystyle \sin x=\sum _{{\text{odd }}k\geq 1}(-1)^{(k-1)/2}\ {x^{k} \over k!}.}
However, a problem arises concerning the interchange in the order of two limiting operations.
cos
x
=
∑
even
k
≥
0
(
−
1
)
k
/
2
x
k
k
!
.
{\displaystyle \cos x=\sum _{{\text{even }}k\geq 0}(-1)^{k/2}\ {x^{k} \over k!}.}
cos
(
∑
i
=
1
∞
θ
i
)
=
∑
even
k
≥
0
(
−
1
)
k
/
2
∑
A
⊆
{
1
,
2
,
3
,
…
}
|
A
|
=
k
(
∏
i
∈
A
sin
θ
i
∏
i
∉
A
cos
θ
i
)
{\displaystyle \cos \left(\sum _{i=1}^{\infty }\theta _{i}\right)=\sum _{{\text{even }}k\geq 0}~(-1)^{k/2}~~\sum _{\begin{smallmatrix}A\subseteq \{\,1,2,3,\dots \,\}\\\left|A\right|=k\end{smallmatrix}}\left(\prod _{i\in A}\sin \theta _{i}\prod _{i\not \in A}\cos \theta _{i}\right)}
![{\displaystyle {\begin{aligned}\sin x&=\sum _{{\text{odd }}k\in \{1,\dots ,n\}}(-1)^{(k-1)/2}{n \choose k}\sin ^{k}\theta \cos ^{n-k}\theta \\[12pt]&=\sum _{{\text{odd }}k\in \{1,\dots ,n\}}\left[(-1)^{(k-1)/2}\left({n(n-1)\cdots (n-k+1) \over k!}\right)\left({x \over n}\right)^{k}\right.\\&{}\qquad \qquad \qquad \qquad \qquad \qquad \qquad \left.{}\cdot \left({\sin(x/n) \over x/n}\right)^{k}\cos ^{n-k}(x/n)\right]\\[12pt]&=\sum _{{\text{odd }}k\in \{1,\dots ,n\}}\left[(-1)^{(k-1)/2}\ {x^{k} \over k!}\cdot {}\right.\\&{}\qquad \qquad \left.{n(n-1)\cdots (n-k+1) \over n^{k}}\cdot \left({\sin(x/n) \over x/n}\right)^{k}\cos ^{n-k}(x/n)\right]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b9a5b497d3434045acfd0f445a3e87a03e0d4721)
