# comparison of $sin\theta $ and $\theta $ near $\theta =0$

###### Theorem 1.

Let $$, where $\theta $ is an angle measured in radians. Then $$.

###### Proof.

Let $O=(0,0)$, $P=(1,0)$, and $Q=(\mathrm{cos}\theta ,\mathrm{sin}\theta )$. Note that the circle ${x}^{2}+{y}^{2}=1$ passes through $P$ and $Q$ and that the shortest arc along this circle from $P$ to $Q$ has length $\theta $. Note also that the line segments^{} $\overline{OP}$ and $\overline{OQ}$ are radii of the circle ${x}^{2}+{y}^{2}=1$ and therefore must each have length 1.

Draw the line segment $\overline{PQ}$. Since this does not correspond to the arc, its length $$.

Drop the perpendicular^{} from $Q$ to $\overline{OP}$. Let $R$ be the point of intersection^{}. Note that $\left|\overline{OR}\right|=\mathrm{cos}\theta $ and $\left|\overline{QR}\right|=\mathrm{sin}\theta $.

Since $$, $$ and $$. Thus, $R\ne Q$ and $R$ lies strictly in between $O$ and $P$. Therefore, $\left|\overline{PR}\right|>0$.

By the Pythagorean theorem^{}, ${\left|\overline{QR}\right|}^{2}+{\left|\overline{PR}\right|}^{2}={\left|\overline{PQ}\right|}^{2}$. Thus, $$. Therefore, $$.
∎

The analogous result for $\theta $ slightly below 0 is:

###### Corollary 1.

Let $$, where $\theta $ is an angle measured in radians. Then $$.

###### Proof.

Since $$, the previous theorem yields $$. Since $\mathrm{sin}$ is an odd function, $$. It follows that $$. ∎

Title | comparison of $sin\theta $ and $\theta $ near $\theta =0$ |
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Canonical name | ComparisonOfsinthetaAndthetaNeartheta0 |

Date of creation | 2013-03-22 16:58:29 |

Last modified on | 2013-03-22 16:58:29 |

Owner | Wkbj79 (1863) |

Last modified by | Wkbj79 (1863) |

Numerical id | 7 |

Author | Wkbj79 (1863) |

Entry type | Theorem |

Classification | msc 26A03 |

Classification | msc 51N20 |

Classification | msc 26A06 |

Related topic | LimitOfDisplaystyleFracsinXxAsXApproaches0 |

Related topic | JordansInequality |