Taylor Series of $sin(x)$ and $cos(x)$
We already know the derivatives of $sin(x)$ and $cos(x)$ and thus we can deduce the following:
$$
\begin{aligned}
{d \over dx}\left( sin(x) \right) & = sin^{(1)}(x) = cos(x)\\
{d \over dx}\left( cos(x) \right) & = cos^{(1)}(x) = sin^{(2)}(x) = -sin(x)\\
{d \over dx}\left( -sin(x) \right) & = cos^{(2)}(x) = sin^{(3)}(x) = -cos(x)\\
{d \over dx}\left( -cos(x) \right) & = cos^{(3)}(x) = sin^{(4)}(x) = sin(x)\\
sin(0) & = 0\\
cos^{(0)}(0) & = sin^{(1)}(0) = cos(0) = 1\\
cos^{(1)}(0) & = sin^{(2)}(0) = -sin(0) = 0\\
cos^{(2)}(0) & = sin^{(3)}(0) = -cos(0) = -1\\
cos^{(3)}(0) & = sin^{(4)}(0) = sin(0) = 0\\
\end{aligned}
$$
We can use these values now to determine the Taylor Series for $cos(x)$ and $sin(x)$:
$$
\begin{aligned}
cos(x) & = \sum_{j=0}^\infty cos^{(j)}(0) \cdot {{x^j} \over j!}\\
& = cos^{(0)}(0) \cdot {{x^0} \over 0!} + cos^{(1)}(0) \cdot {{x^1} \over 1!} + cos^{(2)}(0) \cdot {{x^2} \over 2!} + cos^{(3)}(0) \cdot {{x^3} \over 3!} + ...\\
& = 1 + 0 \cdot x - 1 \cdot {{x^2} \over 2!} - 0 \cdot {{x^3} \over 3!} + 1 \cdot {{x^4} \over 4!} + .....\\
& = 1 - {{x^2} \over 2!} + {{x^4} \over 4!} - ......\\
cos(x) & = \sum_{j=0}^\infty (-1)^j \cdot {{x^{2j}} \over {(2j)!}}\\
sin(x) & = \sum_{j=0}^\infty sin^{(j)}(0) \cdot {{x^j} \over j!}\\
& = sin^{(0)}(0) \cdot {{x^0} \over 0!} + sin^{(1)}(0) \cdot {{x^1} \over 1!} + sin^{(2)}(0) \cdot {{x^2} \over 2!} + sin^{(3)}(0) \cdot {{x^3} \over 3!} + ...\\
& = 0 + 1 \cdot x + 0\cdot {{x^2} \over 2!} - 1 \cdot {{x^3} \over 3!} - 0 \cdot {{x^4} \over 4!} + 1 \cdot {{x^5} \over 5!} + .....\\
& = x - {{x^3} \over 3!} + {{x^5} \over 5!} - ......\\
sin(x) & = \sum_{j=0}^\infty (-1)^j \cdot {{x^{2j+1}} \over {(2j+1)!}}
\end{aligned}
$$
Taylor Series of $tan(x)$
Let's look at determining the Taylor Series for $tan(x)$. Let's start of with determining the derivatives of $tan(x)$ and then the values at $x=0$:
$$
\begin{aligned}
tan^{(1)}(x) & = {d \over dx} \left( tan(x) \right)\\
& = {d \over dx} \left( {{sin(x)} \over {cos(x)}} \right)\\
& = {{cos(x)^2 + sin(x)^2} \over {cos(x)^2}}\\
& = 1 + tan(x)^2\\
tan^{(2)}(x) & = {d \over dx} \left( 1+tan(x)^2 \right) \\
& = 2 \cdot tan(x) \cdot (1 + tan(x)^2)\\
& = 2 \cdot tan(x) + 2 \cdot tan(x)^3\\
tan^{(3)}(x) & = {d \over dx} \left( 2 \cdot tan(x)+ 2 \cdot tan(x)^3 \right)\\
& = 2 \cdot \left(1 + tan(x)^2\right) + 6 \cdot tan(x)^2 \cdot \left( 1 + tan(x)^2 \right)\\
& = 2 + 8 \cdot tan(x)^2 + 6 \cdot tan(x)^4 \\
tan^{(4)}(x) & = {d \over dx} \left( 2 + 8 \cdot tan(x)^2 + 6 \cdot tan(x)^4 \right)\\
& = 16 \cdot tan(x) \cdot \left(1 + tan(x)^2 \right) + 24 \cdot tan(x)^3 \cdot \left( 1 + tan(x)^2 \right)\\
& = 16 \cdot tan(x) + 40 \cdot tan(x)^3 + 24 \cdot tan(x)^5\\
tan^{(5)}(x) & = {d \over dx} \left( 16 \cdot tan(x) + 40 \cdot tan(x)^3 + 24 \cdot tan(x)^5 \right)\\
& = 16 \cdot \left( 1 + tan(x)^2 \right) + 120 \cdot tan(x)^2 \left( 1+tan(x)^2\right) + 120 \cdot tan(x)^4 \left( 1 + tan(x)^2 \right)\\
& = 16 + 136 \cdot tan(x)^2 + 240 \cdot tan(x)^4 + 120 \cdot tan(x)^6\\
\end{aligned}
$$
Now we can determine the values as zero:
$$
\begin{aligned}
tan^{(0)}(0) & = 0\\
tan^{(1)}(0) & = 1 + tan(0)^2 = 1\\
tan^{(2)}(0) & = 2 \cdot tan(0) + 2 \cdot tan(0)^3 = 0\\
tan^{(3)}(0) & = 2 + 8 \cdot tan(0)^2 + 6 \cdot tan(0)^4 = 2\\
tan^{(4)}(0) & = 16 \cdot tan(0) + 40 \cdot tan(0)^3 + 24 \cdot tan(0)^5 = 0\\
tan^{(5)}(0) & = 16 + 136 \cdot tan(0)^2 + 240 \cdot tan(0)^4 + 120 \cdot tan(0)^6 = 16\\
tan^{(6)}(0) & = 0\\
tan^{(7)}(0) & = 272\\
tan^{(8)}(0) & = 0\\
tan^{(9)}(0) & = 7936\\
tan^{(10)}(0) & = 0\\
tan^{(11)}(0) & = 353792\\
tan(x) & = 0 + x + 0 + 2 \cdot {{x^3} \over 6} + 0 + 16 \cdot {{x^5} \over 120} + 0 + 272 \cdot {{x^7} \over 5040} + 0 + 7936 \cdot {{x^9} \over 362880} + 0 + 353792 \cdot {{x^9} \over 39916800} + 0 + .....\\
& = x + {1 \over 3}{x^3} + {2 \over 15}x^5 + {17 \over 315}x^7 + {62 \over 2835}x^9 + {1382 \over 155925}x^{11}+ ...
\end{aligned}
$$