Sine of Zero is Zero
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Theorem
- $\sin 0 = 0$
where $\sin$ denotes the sine.
Proof
Recall the definition of the sine function:
- $\ds \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$
Thus:
- $\ds \sin 0 = 0 - \frac {0^3} {3!} + \frac {0^5} {5!} - \cdots = 0$
$\blacksquare$
Also see
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Special angles
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 16.3 \ (1) \ \text{(ii)}$
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): Appendix $12$: Trigonometric formulae: Trigonometric values for some special angles
- 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): Appendix $14$: Trigonometric formulae: Trigonometric values for some special angles