1. **State the problem:** We want to show that the functions defined by the series $$aR_{2,0}(-a^2,0,t) = a \sum_{n=0}^\infty \frac{(-a^2)^n t^{(n+1)2-1}}{(2n+1)!}$$ and $$R_{2,1}(-a^2,0,t) = \sum_{n=0}^\infty \frac{(-a^2)^n t^{(n+1)2-1-1}}{\Gamma((n+1)2-1)}$$ are equivalent to the sine and cosine functions respectively. 2. **Analyze the sine function series:** Rewrite the exponent and factorial in the sine series: $$t^{(n+1)2-1} = t^{2n+1}$$ and the denominator is $(2n+1)!$. So the series becomes: $$a \sum_{n=0}^\infty \frac{(-a^2)^n t^{2n+1}}{(2n+1)!} = a \left(t - \frac{a^2 t^3}{3!} + \frac{a^4 t^5}{5!} - \ldots \right)$$ This matches the Taylor series expansion of $\sin(at)$: $$\sin(at) = \sum_{n=0}^\infty (-1)^n \frac{(at)^{2n+1}}{(2n+1)!} = a \left(t - \frac{a^2 t^3}{3!} + \frac{a^4 t^5}{5!} - \ldots \right)$$ 3. **Conclusion for sine:** Therefore, $$aR_{2,0}(-a^2,0,t) = \sin(at).$$ 4. **Analyze the cosine function series:** Rewrite the exponent and gamma function in the cosine series: $$t^{(n+1)2-1-1} = t^{2n}$$ and $$\Gamma((n+1)2-1) = \Gamma(2n+1) = (2n)!$$ So the series becomes: $$\sum_{n=0}^\infty \frac{(-a^2)^n t^{2n}}{(2n)!} = 1 - \frac{a^2 t^2}{2!} + \frac{a^4 t^4}{4!} - \ldots$$ This matches the Taylor series expansion of $\cos(at)$: $$\cos(at) = \sum_{n=0}^\infty (-1)^n \frac{(at)^{2n}}{(2n)!} = 1 - \frac{a^2 t^2}{2!} + \frac{a^4 t^4}{4!} - \ldots$$ 5. **Conclusion for cosine:** Therefore, $$R_{2,1}(-a^2,0,t) = \cos(at).$$ **Final boxed results:** $$\boxed{aR_{2,0}(-a^2,0,t) = \sin(at)}$$ $$\boxed{R_{2,1}(-a^2,0,t) = \cos(at)}$$