We consider the
cosine series
-
![{\displaystyle {}\cos x=\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n}}{(2n)!}}\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/66458c604ba595622175e1e2e96a47f1c1e3567f)
For
,
we have
.
For
,
one can write
![{\displaystyle {}{\begin{aligned}\cos 2&=1-{\frac {2^{2}}{2!}}+{\frac {2^{4}}{4!}}-{\frac {2^{6}}{6!}}+{\frac {2^{8}}{8!}}-\ldots \\&=1-{\frac {2^{2}}{2!}}{\left(1-{\frac {4}{3\cdot 4}}\right)}-{\frac {2^{6}}{6!}}{\left(1-{\frac {4}{7\cdot 8}}\right)}-\ldots \\&=1-2(2/3)-\ldots \\&\leq -1/3.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/843136344ce66cd07fd8dda5dc411cee1f16e2ca)
Hence, due to
the intermediate value theorem,
there exists at least one zero in the given interval.
To prove uniqueness, we consider the
derivative
of cosine, which is
-
![{\displaystyle {}\cos 'x=-\sin x\,,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6eec6685338ffde36b4977be8417d5a002c5c7cd)
due to
fact.
Hence, it is enough to show that sine is positive in the interval
, because then cosine is
strictly decreasing
by
fact
in the interval and there is only one zero. Now, for
,
we have
![{\displaystyle {}{\begin{aligned}\sin x&=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\ldots \\&=x{\left(1-{\frac {x^{2}}{3!}}\right)}+{\frac {x^{5}}{5!}}{\left(1-{\frac {x^{2}}{6\cdot 7}}\right)}+\ldots \\&\geq x{\left(1-{\frac {4}{3!}}\right)}+{\frac {x^{5}}{5!}}{\left(1-{\frac {4}{6\cdot 7}}\right)}+\ldots \\&\geq x/3\\&>0.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1db067b6d0b6d682255b563ede8cc5a1110e0f6)