It is not hard to show that we can modify the functional equation of the riemann $\zeta$-function to
$$\zeta(1-s)=2^{1-s}\pi^{-s}\cos\frac{\pi s}{2} \Gamma(s)\zeta(s)$$
Now the textbook claims that $$2^{1-s}\pi^{-s}\cos\frac{\pi s}{2} \Gamma(s)=O(\vert t\vert^{\frac{1}{2}-\sigma})$$ for $-5\leq\sigma\leq5$, $s=\sigma+it$
The first two factors are supposed to be bounded on this and are no problem. But I am not sure how to deal with the cosine and the $\Gamma$-function. I thought about applying Stirlings formula, but it is only available for $\sigma>0$. Also I dont know how to deal with the cosine.
Maybe I am lagging some standard knowledge about the $\Gamma$-function and there are some supposedly well known estimations.
The author does not lose any words about how this is supposed to be proven so I would appreciate any help or hints how to deal with this problem
To understand the cosine term, write it as a sum of exponentials.
For $\Gamma$, there is a well-known form of Stirling's formula that is uniform approximately in wedges away from the negative real axis and $0$. Stated more rigorously, the following is true.
In the fixed region that you're considering, where it's not necessary to have large negative real parts, it would also be sufficient to use a Stirling estimate for $\mathrm{Re}\, s > 0$ and the functional equation for $\Gamma$ to translate this estimate to an estimate for $\mathrm{Re}\, s > -5$, say.
In practice, a very useful form of Stirling's Formula is the following:
$$\begin{align} \lvert \Gamma(z) \rvert &\asymp e^{- \mathrm{Re} z} \lvert z \rvert^{\mathrm{Re} z - \frac{1}{2}} e^{- (\arg z)(\mathrm{Im} z)} \tag{1}\\ &\asymp \lvert t \rvert^{\sigma - \frac{1}{2}} e^{- \frac{\pi t}{2} } \tag{2}, \end{align}$$ where the second line writes $z = \sigma + it$ and holds for $\sigma$ constrained to a fixed vertical strip as $\lvert t \rvert \to \infty$. This last line is often sufficient for number theoretical bounding (including your question).
Let us first consider your question with this form, and then show how to get this form of Stirling from the previous (more commonly stated) version.
You consider $$ \zeta(1-s)=2^{1-s}\pi^{-s}\cos\frac{\pi s}{2} \Gamma(s)\zeta(s). $$ I'll write $s = \sigma + it$ now. For $\sigma$ in any fixed strip and as $\lvert t \rvert \to \infty$, the factors $2^{1 - s}$ and $\pi^{-s}$ are absolutely bounded in size. I ignore them.
Write $\cos \frac{\pi s}{2} = (1/2)(e^{i \pi s /2} + e^{-i \pi s / 2}),$ so that $\cos \frac{\pi s}{2} = O\big( e^{\pi \lvert t \rvert / 2} \big)$ in strips. This exponential cancels with the exponential in the final approximation in the last form of Stirling's that I gave above, leaving just $$ 2^{1-s}\pi^{-s}\cos\frac{\pi s}{2} \Gamma(s) \ll \lvert t \rvert^{\sigma - (1/2)}. $$ I note that the exponent is the negative of the inverse in your original question, but this one is correct.
To prove this absolute value form of Stirling, let us continue to write $z = \sigma + it$ and discard $\sqrt{2\pi}$ immediately. Then we have that $$ \lvert \Gamma(z) \rvert \asymp \lvert e^{-z} \rvert \, \lvert z^{z - (1/2)} \rvert. $$ The exponential is easy. It appears in $(1)$ and is uniformly bounded in vertical strips, so it doesn't need to appear in $(2)$. We now use that $$ z^{z - (1/2)} = e^{(z - 1/2)(\log \lvert z \rvert + i \arg z)} = \lvert z \rvert^{z - (1/2)} e^{(z - 1/2)(i \arg z)}. $$ Taking absolute values, we find that $$ \lvert z^{z - (1/2)} \rvert = \lvert z \rvert^{\sigma - (1/2)} e^{- t \arg z }. $$ This proves $(1)$.
To prove $(2)$, we fix $\sigma$ to a strip and think of $t$ as being large. Then $\lvert z \rvert \approx \lvert t \rvert$, and $\arg z \to \pm \pi/2$. One can show that these approximations are correct to first order (though like many Stirling bounds, this is annoying). Note that the sign of $t$ and $\arg z$ agree, which is why the exponential becomes $e^{- \pi \lvert t \rvert / 2 }$, ignoring the sign of $t$. This (mostly) proves $(2)$.