Sine Inequality/Proof 2
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Theorem
- $\size {\sin x} \le \size x$
for all $x \in \R$.
Proof
For $x = 0$, the inequality is trivial, as $\sin 0 = 0$.
If $\size x \ge 1$, the inequality also clear as:
| \(\ds \size {\sin x}\) | \(\le\) | \(\ds 1\) | Real Sine Function is Bounded | |||||||||||
| \(\ds \) | \(\le\) | \(\ds \size x\) |
Now, suppose $0 < \size x < 1$
Then on the one hand:
| \(\ds \dfrac {\sin x} x\) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n + 1}!}\) | Definition of Real Sine Function | |||||||||||
| \(\ds \) | \(=\) | \(\ds 1 - \sum_{n \mathop = 1}^\infty \paren {-1}^{n-1} \frac {x^{2 n} } {\paren {2 n + 1}!}\) | ||||||||||||
| \(\ds \) | \(\ge\) | \(\ds 1 - \sum_{k \mathop = 0}^\infty \frac {x^{4k+2} } {\paren {4k+3}!}\) | as $x^2 \ge 0$ | |||||||||||
| \(\ds \) | \(\ge\) | \(\ds 1 - \sum_{k \mathop = 0}^\infty \frac 1 {\paren {4k+3}!}\) | as $x^2 \le 1$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds 1 - \frac 1 {3!} \paren {1 + \frac 1 {7 \cdot 6 \cdot 5 \cdot 4} + \frac 1 {11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4} + \cdots }\) | ||||||||||||
| \(\ds \) | \(\ge\) | \(\ds 1 - \frac 1 {3!} \paren {1 + \frac 1 {4^4} + \frac 1 {4^8} + \frac 1 {4^{12} } + \cdots }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1 - \frac {4^4} {3! \paren {4^4 - 1} }\) | Sum of Infinite Geometric Sequence | |||||||||||
| \(\ds \) | \(\ge\) | \(\ds 0\) |
On the one hand:
| \(\ds \dfrac {\sin x} x\) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n + 1}!}\) | Definition of Real Sine Function | |||||||||||
| \(\ds \) | \(=\) | \(\ds 1 - \dfrac {x^2} 6 + \sum_{n \mathop = 2}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n + 1}!}\) | ||||||||||||
| \(\ds \) | \(\le\) | \(\ds 1 - \dfrac {x^2} 6 + \sum_{k \mathop = 1}^\infty \frac {x^{4k} } {\paren {4k+1}!}\) | as $x^2 \ge 0$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds 1 - x^2 \paren {\frac 1 6 - \sum_{k \mathop = 1}^\infty \frac {x^{2 \paren {k-1} } } {\paren {4k+1}!} }\) | ||||||||||||
| \(\ds \) | \(\le\) | \(\ds 1 - x^2 \paren {\dfrac 1 6 - \sum_{k \mathop = 1}^\infty \frac 1 {\paren {4 k + 1}!} }\) | as $x^2 \le 1$ |
The last expression is $\le 1$, since:
| \(\ds \sum_{k \mathop = 1}^\infty \frac 1 {\paren {4 k + 1}!}\) | \(=\) | \(\ds \frac 1 {5!} \sum_{k \mathop = 1}^\infty \frac {5 !} {\paren {4 k + 1}!}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {5!} \paren {1 + \frac 1 {9 \cdot 8 \cdot 7 \cdot 6} + \frac 1 {13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6} + \cdots}\) | ||||||||||||
| \(\ds \) | \(\le\) | \(\ds \frac 1 {5!} \paren {1 + \frac 1 {6^4} + \frac 1 {6^8} + \frac 1 {6^{12} } + \cdots}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {5!} \frac {6^4} {6^4 - 1}\) | Sum of Infinite Geometric Sequence | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 6 \frac {6^4} {20 \paren {6^4 - 1} }\) | ||||||||||||
| \(\ds \) | \(<\) | \(\ds \frac 1 6\) |
$\blacksquare$