spreads the result to the case of $q$-Sumudu transform giving $$S_{q}\left( e_{q}\right) \left( ax\right) \left( u\right) =\frac{1}{1-au},\text{ }\left\vert au\right\vert <1.$$Similarly, by fixing $u=1$ and consulting $\left( 20\right) $ yield the following case of $q$-Laplace transform $$L_{q}\left( e_{q}\left( ax\right) \right) \left( v\right) =L_{q}\left(
e_{q}\left( ax\right) \right) \left( v\right) =\frac{1}{v-a},\text{ }\left\vert a\right\vert <v.$$From above investigations, we, further, deduce$$N_{q}\left( \sin _{q}ax\right) \left( u;v\right) =N_{q}\left( \dfrac{e_{q}\left( iax\right) -e_{q}\left( -iax\right) }{2i}\right) \left(
u;v\right) =\frac{au}{v^{2}+a^{2}u^{2}},\text{ }\left\vert au\right\vert <v,
\tag{35}$$

and$$N_{q}\left( \cos _{q}ax\right) \left( u;v\right) =N_{q}\left( \dfrac{e_{q}\left( iax\right) +e_{q}\left( -iax\right) }{2i}\right) \left(
u;v\right) =\frac{v}{v^{2}+a^{2}u^{2}},\text{ }\left\vert au\right\vert <v. 
\tag{36}$$Hence, by virtue of $\left( 35\right) $ and $\left( 36\right) $ we state without proof the following corrollary.

<span style="font-variant:small-caps;">Corrollary 4.</span> *Let* $a$* be a real number. Then, the following hold.*

$\left( \text{i}\right) S_{q}\left( \sin _{q}ax\right) \left( u\right) =\dfrac{au}{1+a^{2}u^{2}},$ $\left\vert au\right\vert <1;$ $\ \ \ \left( 
\text{ii}\right) S_{q}\left( \cos _{q}ax\right) \left( u\right) =\dfrac{1}{1+a^{2}u^{2}},$ $\left\vert au\right\vert <1,$

$\left( \text{iii}\right) L_{q}\left( \sin _{q}ax\right) \left( v\right) =\dfrac{a}{v^{2}+a^{2}},$ $\left\vert a\right\vert <v;$ $\ \ \ \ \ \left( 
\text{iv}\right) L_{q}\left( \cos _{q}ax\right) \left( v\right) =\dfrac{v}{v^{2}+a^{2}},$ $\left\vert a\right\vert <v.$

From Theorem 1 we state and prove the following corrollary.

<span style="font-variant:small-caps;">Corrollary 5.</span> *Let* $a$* be a real number and* $f\left( x\right) =\dsum\limits_{n\geq 0}A_{n}x^{n}$*. Then, we*$\ $*have 