Pauli Matrices proof The original question was to prove

$$e^{i\sigma_z\phi} \sigma_y e^{-i\sigma_z\phi}=e^{2i\sigma_z\phi} \sigma_y.$$

Since left hand side is equal to $\sigma_y$, I thought in order for right hand side to be equal to $\sigma_y$,I have to show that $$e^{2i\sigma_z\phi}=1$$
For this,I tried to expand:
$$e^{2i\sigma_z\phi}=\cos(2\sigma_z\phi)+i\sin(2\sigma_z\phi)$$
And use, 
$$\cos(2\sigma_z\phi)=1-\frac{(2\sigma_z\phi)^2}{2!}+\frac{(2\sigma_z\phi)^4}{4!}....$$
$$\sin(2\sigma_z\phi)=2\sigma_z\phi-\frac{(2\sigma_z\phi)^3}{3!}+\frac{(\sigma_2\phi)^5}{5!}...$$
where $\sigma_z$ is a Pauli matrix ($z$-component)
I have tried to expand the exponential in the similar way as above, but could not verify the above result. Don't know, whether my question is wrong or there are any other methods to solve this problem.
 A: Use fact that the matrices $i\,\sigma$ form the Lie algebra $\mathfrak{su}(2)$ and then the adjoint representation "braiding relationship" that:
$$\exp(\phi\,\mathrm{ad}(Z))\,Y = \mathrm{Ad}(e^{\phi\,Z})\,Y\tag{1}$$
and here $Z$ and $Y$ are $3\times 1$ column matrices that stand for superpositions of basis matrices in the Lie algebra in question. $e^{i\sigma_z\phi}.\sigma_y.e^{-i\sigma_z\phi}$ is the linear map $\mathrm{Ad}(e^{i\,\phi\,\sigma_z})$ acting on the vector that stands for $i\,\sigma_y$.
Now, $\mathrm{ad}(Z)\,Y$ stands for the commutator bracket $[i\,\sigma_z,\,i\,\sigma_y] = -[\sigma_z,\,\sigma_y] = 2\,i\,\sigma_x$. Thus $\mathrm{ad}(Z)^2\,Y$ stands for the commutator bracket $[i\,\sigma_z,\,2\,i\,\sigma_x] = -2\,[\sigma_z,\,\sigma_x] = -4\,i\,\sigma_y$. Keep on repeating these commutator brackettings to find the LHS of (1), which is the superposition:
$$\begin{array}{lcl}\exp(\phi\,\mathrm{ad}(Z))\,Y &=& Y + 2 \,\phi\,X - \frac{(2\,\phi)^2}{2!}\,Y -  \frac{(2\,\phi)^3}{3!}\,X +  \frac{(2\,\phi)^4}{4!}\,Y + \frac{(2\,\phi)^5}{5!}\,X -\cdots\\\\&=& \cos(2\,\phi)\,Y + \sin(2\,\phi)\,X\end{array}\tag{2}$$
which Pauli matrix does this superposition stand for?

More First Principles Answer
Actually you use a generalized version of something very like this procedure to prove the braiding relationship above.
Differentiate the LHS wrt $\phi$ to find:
$$\mathrm{d}_\phi e^{i\sigma_z\phi}.\sigma_y.e^{-i\sigma_z\phi} = i\,e^{i\sigma_z\phi}.[\sigma_z,\,\sigma_y].e^{-i\sigma_z\phi} = i\,[\sigma_z,\,e^{i\sigma_z\phi}.\sigma_y.e^{-i\sigma_z\phi}]\tag{3}$$
Use the commutation relationships to prove that this function $f(\phi)$ of $\phi$ equals $2\,\sigma_x$ at $\phi=0$. 
Do the same for the RHS to prove: 
$$\mathrm{d}_\phi e^{2i\sigma_z\phi}.\sigma_y = 2\,i\,e^{2i\sigma_z\phi}.\sigma_z.\sigma_y\tag{4}$$
A function $g(\phi)$ of $\phi$ which also equals $2\,\sigma_x$ at $\phi=0$ (prove this by direct evaluation).
Now find a differential equation that describes $f$ in terms of its derivatives alone. Do likewise for $g$. Then prove that the two sets are the same Cauchy initial value problem, thus by uniqueness of solution (Picard-Lindelöf theorem) $f(\phi) = g(\phi)$.
A: LHS: $(\cos\phi+i\sigma_z\sin\phi)\sigma_y(\cos\phi-i\sigma_z\sin\phi)=\cos^2\phi\sigma_y-i\cos\phi\sin\phi[\sigma_y,\sigma_z]+\sin^2\phi \sigma_z\sigma_y\sigma_z=(\cos^2\phi-\sin^2\phi)\sigma_y+2\cos\phi\sin\phi\sigma_x=\cos 2\phi \sigma_y+\sigma_x\sin2\phi=(\cos 2\phi+i\sin 2\phi\sigma_z)\sigma_y=e^{2i\phi\sigma_z}\sigma_y$
A: Here is a some different proof:
\begin{align*}
e^{i\sigma_{z}\phi}\sigma_ye^{-i\sigma_z\phi}
&=e^{i\sigma_{z}\phi}(e^{-i\sigma_z\phi}\sigma_y+[\sigma_y,e^{-i\sigma_z\phi}])\\
&=\sigma_y+e^{i\sigma_{z}\phi}[\sigma_y,cos(\phi)-i sin(\sigma_z\phi)]\\
&=\sigma_y+e^{i\sigma_{z}\phi}[\sigma_y,-i\sigma_zsin(\phi)]\\
&=\sigma_y-i sin(\phi) e^{i\sigma_{z}\phi}[\sigma_y,\sigma_z]\\
&=\sigma_y+2isin(\phi) e^{i\sigma_z\phi}\sigma_z\sigma_y\\
&=e^{i\sigma_z\phi}(e^{-i\sigma_z\phi}\sigma_y+2i\sigma_zsin(\phi)\sigma_z\sigma_y)\\
&=e^{i\sigma_z\phi}(cos(\sigma_z\phi)+i\sigma_zsin(\phi))\sigma_y\\
&=e^{2i\sigma_z\phi}\sigma_y
\end{align*}
