Why oxidation potential of an electrode equals negative reduction potential of the same electrode?
There exist two methods for teaching electrochemistry, and of course two schools of teachers. The two methods are equivalent, but they are exactly the opposite of one another. I will take an example to explain it correctly : Let us speak of the Daniell cell Zn/Cu, which uses reduction potentials equal to -$0.76 $ V for Zn and +$0.34$ V for Cu, if the ions concentrations are 1 molar. In this case, the overall voltage is $1.10$ V.
The Daniell cell is based on the two half-reactions :  $$\ce{Zn^{2+} + 2e^- -&gt; Zn}$$  $$\ce{Cu^{2+} +2e^--&gt; Cu}$$In the Zn/Cu cell, Zn is oxidized and acts as emitter of electron, which is in the opposite direction of the above equation, so the equation explaining what is going on in a Daniell cell contains the equation of Zn reversed. This becomes :$$\ce{Zn-&gt;  Zn^{2+} + 2e^-}$$  $$\ce{Cu^{2+} + 2e- -&gt;Cu}$$When you want to describe the overall equation that goes in this cell, you add these two half-equations, remove the 2$\ e^-$ and this gives : $$\ce{Zn + Cu^{2+}-&gt;  Zn^{2+} + Cu}$$  We are here at the point where the two approaches differ. Strangely enough, one uses an addition, and the second a subtraction. Let me speak of the first method, which is used in your textbook. 

First method. To obtain this overall equation, you had to add the two half-equations. To follow the same logic, you may simultaneously add the corresponding reduction potentials. But the reduction potentials of Zn and Cu are respectively $-0.76$V, and + $0.34$ V. And if you add them, you do not get $1.10$ V. So the chemists of this school have decided that when a half- equation is used in the opposite sense, with the electrons at the right-hand side, in an oxidation sense, like Zn here, the sign of the potential of this cell should also be opposite. $\ E_{anode} = - E_{cathode}$. Here : $E(Zn)_{anode} = - E(Zn)_{cathode)}$ = +$0.76$ V. With this assumption, you add $+ 0.76$ V plus $+0.34$ V and you obtain $1.10$ V, which is perfect. 

Second method. Here you use the same half-reactions as described previously. But the values of the potential remains the same if the electrode works as a reducing agent or as an oxidant. It is always - $0.76$ V pour Zn and + $0.34$ V for Cu. You never change its sign. But remembering from the electricity course, that a measured voltage is a difference of potential, the voltage of the cell may be obtained by subtracting the redox potential of the cathode minus the potential of the anode, that is $E_{Cu} - E_{Zn}$. The result of this subtraction is: $+0.34$V -($0.76$ V) = $1.10$ V, which is perfect. This remains true even though you have obtained the overall equation with an addition of two half-reactions.

As you see, both approaches are equivalent. Have you understood my development ?