Why: $z^{-1}=\cos (-\phi)+i\sin(-\phi)$ Let's suppose that we have a complex number with
$$r=1 \implies z=\cos\phi+i\sin\phi$$
Then why is $$z^{-1} =\cos (-\phi)+i\sin (-\phi)=\cos\phi-i\sin\phi$$

As your complex number as $r=1$, you can express it like $z=e^{i\theta}$, where $\theta$ is the argument. Then,
$$z^{-1}=\frac{1}{z}=\frac{1}{e^{i\theta}}=e^{-i\theta}$$
Now, using the trigonometric form of complex numbers, 
$$e^{-i\theta}=\cos(-\theta)+i\sin(-\theta)=\cos(\theta)-i\sin(\theta),$$
where we used that $\cos(\theta)=\cos(-\theta)$ and $\sin(\theta)=-\sin(-\theta)$ for all $\theta$. 
Edit: 
The power series of the exponential, the sine and the cosine are 
$$e^x=\sum_{n=0}^\infty \dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$$
$$\sin x=\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!}x^{2n+1}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+\dfrac{x^9}{9!}+...$$
$$\cos x=\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n)!}x^{2n}=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+\dfrac{x^8}{8!}+...$$
Now, substituting $i\theta$ instead of $x$ in the power series of $e^x$, we have
\begin{align*}
e^{i\theta} & =1+i\theta+\dfrac{(i\theta)^2}{2!}+\dfrac{(i\theta)^3}{3!}+\dfrac{(i\theta)^4}{4!}+...=1+i\theta+\dfrac{i^2\theta^2}{2!}+\dfrac{i^3\theta^3}{3!}+\dfrac{i^4\theta^4}{4!}+... \\ \\
& =1+i\theta-\dfrac{\theta^2}{2!}-\dfrac{i\theta^3}{3!}+\dfrac{\theta^4}{4!}+...=\left(1-\dfrac{\theta^2}{2!}+\dfrac{\theta^4}{4!}-...\right)+i\left(\theta-\dfrac{i\theta^3}{3!}+\dfrac{\theta^5}{5!}-...\right) \\ \\
& =\cos\theta +i\sin\theta
\end{align*}
