Cosine of 2 is Strictly Negative

Theorem

$\cos 2 < 0$

where $\cos$ denotes the cosine.

Proof

Recall the definition of the cosine:

The real function $\cos: \R \to \R$ is defined as:

\(\ds \cos x\)

\(=\)

\(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n!} }\)

\(\ds \)

\(=\)

\(\ds 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \frac {x^6} {6!} + \cdots + \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} + \cdots\)

Thus:

\(\ds \cos 2\)

\(=\)

\(\ds 1 - \paren {\frac {2^2} {2!} - \frac {2^4} {4!} } - \paren {\frac {2^6} {6!} - \frac {2^8} {8!} } - \cdots\)

\(\ds \)

\(=\)

\(\ds 1 - \sum_{n \mathop = 1}^\infty A_n\)

where:

\(\ds A_n\)

\(:=\)

\(\ds \dfrac {2^{4 n - 2} } {\paren {4 n - 2}!} - \dfrac {2^{4 n} } {\paren {4 n}!}\)

\(\ds \)

\(=\)

\(\ds \dfrac {4 n \paren {4n - 1} 2^{4 n - 2} - 2^{4 n} } {\paren {4 n}!}\)

\(\ds \)

\(=\)

\(\ds \dfrac {n \paren {4n - 2} 2^{4 n} } {\paren {4 n}!}\)

\(\ds \)

\(\ge\)

\(\ds 0\)

for all $n \ge 1$

Thus:

\(\ds \cos 2\)

\(\le\)

\(\ds 1 - A_1\)

\(\ds \)

\(=\)

\(\ds 1 - \paren {\frac {2^2} {2!} - \frac {2^4} {4!} }\)

\(\ds \)

\(=\)

\(\ds - \dfrac 1 3\)

\(\ds \)

\(<\)

\(\ds 0\)

$\blacksquare$