The FunctionAdvisor(identities, math_function) command returns a list of identities for that function.

$\mathrm{FunctionAdvisor}\left(\mathrm{identities}\,\sin \right)$

$\left[\sin \left(\arcsin \left(z\right)\right)=z\,\sin \left(z\right)=-\sin \left(-z\right)\,\sin \left(z\right)=2\sin \left(\frac{z}{2}\right)\cos \left(\frac{z}{2}\right)\,\sin \left(z\right)=\frac{1}{\csc \left(z\right)}\,\sin \left(z\right)=\frac{2\tan \left(\frac{z}{2}\right)}{1+{\tan \left(\frac{z}{2}\right)}^2}\,\sin \left(z\right)=-\frac{\mathrm{I}}{2}\left({\ⅇ}^{\mathrm{I}z}-{\ⅇ}^{\mathrm{-I}z}\right)\,{\sin \left(z\right)}^2=1-{\cos \left(z\right)}^2\,{\sin \left(z\right)}^2=\frac{1}{2}-\frac{\cos \left(2z\right)}{2}\right]$

$\mathrm{FunctionAdvisor}\left(\mathrm{describe}\,\coth \right)$

$\coth =\mathrm{hyperbolic\; cotangent}$

$\mathrm{FunctionAdvisor}\left(\mathrm{definition}\,\coth \right)$

$\left[\coth \left(z\right)=\frac{{\left({\ⅇ}^z\right)}^2+1}{{\left({\ⅇ}^z\right)}^2-1}\,\mathrm{with\; no\; restrictions\; on}\left(z\right)\right]$

$\mathrm{FunctionAdvisor}\left(\mathrm{identities}\,\coth \right)$

$\left[\coth \left(z\right)=-\coth \left(-z\right)\,\coth \left(z\right)=-\frac{-{\coth \left(\frac{z}{2}\right)}^2-1}{2\coth \left(\frac{z}{2}\right)}\,\coth \left(z\right)=\frac{\cosh \left(z\right)}{\sinh \left(z\right)}\,\coth \left(z\right)=\frac{1+\cosh \left(2z\right)}{\sinh \left(2z\right)}\,\coth \left(z\right)=-\frac{\sinh \left(2z\right)}{1-\cosh \left(2z\right)}\,\coth \left(z\right)=\frac{1}{\tanh \left(z\right)}\,\coth \left(z\right)=\frac{1+{\tanh \left(\frac{z}{2}\right)}^2}{2\tanh \left(\frac{z}{2}\right)}\,\coth \left(z\right)=\mathrm{I}\left(-\frac{\mathrm{I}\coth \left(\frac{z}{2}\right)}{2}-\frac{\mathrm{I}\tanh \left(\frac{z}{2}\right)}{2}\right)\,\coth \left(z\right)=-\frac{{\ⅇ}^{-z}+{\ⅇ}^z}{{\ⅇ}^{-z}-{\ⅇ}^z}\,\coth \left(z\right)=\mathrm{I}\left(-\mathrm{I}\mathrm{csch}\left(2z\right)-\mathrm{I}\coth \left(2z\right)\right)\right]$

$\mathrm{eq1}≔\mathrm{FunctionAdvisor}\left(\mathrm{identities}\,\mathrm{BesselI}\right)$

$\mathrm{eq1}≔\left[\mathrm{BesselI}\left(a\,-z\right)=\frac{{\left(-z\right)}^a\mathrm{BesselI}\left(a\,z\right)}{z^a}\,\mathrm{BesselI}\left(a\,\mathrm{I}z\right)=\frac{{\left(\mathrm{I}z\right)}^a\mathrm{BesselJ}\left(a\,z\right)}{z^a}\,\mathrm{BesselI}\left(a+1\,z\right)\mathrm{BesselI}\left(-a\,z\right)-\mathrm{BesselI}\left(-1-a\,z\right)\mathrm{BesselI}\left(a\,z\right)=\frac{2\sin \left(\mathrm{\pi }a\right)}{\mathrm{\pi }z}\,\mathrm{BesselI}\left(a\,\sqrt{z^2}\right)=\frac{{\left(z^2\right)}^{\frac{a}{2}}\mathrm{BesselI}\left(a\,z\right)}{z^a}\,\left[\mathrm{BesselI}\left(a\,b{\left(c{z}^q\right)}^p\right)=\frac{{\left(b{\left(c{z}^q\right)}^p\right)}^a\mathrm{BesselI}\left(a\,b{c}^p{z}^{pq}\right)}{{\left(b{c}^p{z}^{pq}\right)}^a}\,\left(2p\right)::\mathbb{Z}\right]\,\mathrm{BesselI}\left(a\,z\right)=-\frac{2\left(a-1\right)\mathrm{BesselI}\left(a-1\,z\right)}{z}+\mathrm{BesselI}\left(a-2\,z\right)\,\mathrm{BesselI}\left(a\,z\right)=\frac{2\left(a+1\right)\mathrm{BesselI}\left(a+1\,z\right)}{z}+\mathrm{BesselI}\left(a+2\,z\right)\right]$

$\mathrm{depends}\left(\left[\mathrm{eq1}\right]\,a\right),\mathrm{depends}\left(\left[\mathrm{eq1}\right]\,z\right)$

$\mathrm{false},\mathrm{false}$

$\mathrm{eq2}≔\mathrm{FunctionAdvisor}\left(\mathrm{identities}\,\mathrm{Ei}\left(a\,z\right)\right)$

$\mathrm{eq2}≔\left[{\mathrm{Ei}}_1\left(z\right)=-\mathrm{Ei}\left(-z\right)+\frac{\ln \left(-z\right)}{2}-\frac{\ln \left(-\frac{1}{z}\right)}{2}-\ln \left(z\right)\,\left[{\mathrm{Ei}}_{-a}\left(z\right)=a!{\ⅇ}^{-z}\left(\sum _{\mathrm{\_k1}=0}^a\frac{z^{\mathrm{\_k1}-a-1}}{\mathrm{\_k1}!}\right)\,a::{\mathbb{Z}}^{\left(0\,+\right)}\right]\,{\mathrm{Ei}}_a\left(z\right)=\frac{z{\mathrm{Ei}}_{-2+a}\left(z\right)+\left(-2+a-z\right){\mathrm{Ei}}_{-1+a}\left(z\right)}{-1+a}\,{\mathrm{Ei}}_a\left(z\right)=\frac{\left(-a+z\right){\mathrm{Ei}}_{1+a}\left(z\right)+\left(1+a\right){\mathrm{Ei}}_{2+a}\left(z\right)}{z}\right]$

$\mathrm{depends}\left(\left[\mathrm{eq2}\right]\,a\right),\mathrm{depends}\left(\left[\mathrm{eq2}\right]\,z\right)$

$\mathrm{true},\mathrm{true}$