Sine of Zero is Zero

Theorem

$\sin 0 = 0$

where $\sin 0$ is the sine of $0$.

Proof

Recall the definition of the sine function:

$\displaystyle \sin x = \sum_{n=0}^\infty \left({-1}\right)^n \frac {x^{2n+1}}{\left({2n+1}\right)!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$

Thus:

$\displaystyle \sin 0 = 0 - \frac {0^3} {3!} + \frac {0^5} {5!} - \cdots = 0$

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 16.3 \ (1) \ \text{(ii)}$