Sine

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The sine function $\sin \colon \mathbb{R} \rightarrow \mathbb{R}$ is the unique solution of the second order initial value problem $y=-y;y(0)=0,y'(0)=1$.

Properties

Theorem

The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \sin(z) = \cos(z),$$ where $\sin$ denotes the sine function and $\cos$ denotes the cosine function.

Proof

From the definition, $$\sin(z) = \dfrac{e^{iz}-e^{-iz}}{2i},$$ and so using the derivative of the exponential function, the linear property of the derivative, the chain rule, and the definition of the cosine function, $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}z} \sin(z) &= \dfrac{1}{2i} \left[ \dfrac{\mathrm{d}}{\mathrm{d}z} [e^{iz}] - \dfrac{\mathrm{d}}{\mathrm{d}z}[e^{-iz}] \right] \\ &= \dfrac{1}{2i} \left[ ie^{iz} + ie^{-iz} \right] \\ &= \dfrac{e^{iz}+e^{-iz}}{2} \\ &= \cos(z), \end{array}$$ as was to be shown. █

References

Proposition: $\sin(x)=\displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^kx^{2k+1}}{(2k+1)!}$

Proof: proof goes here █

Proposition: $\sin(x) = x \displaystyle\prod_{k=1}^{\infty} \left( 1 - \dfrac{x^2}{k^2\pi^2} \right)$

Proof: proof goes here █

  1. REDIRECT Gamma(z)Gamma(1-z)=pi/sin(pi z)

References

The sine product formula and the gamma function