From the graph of $D_f(\alpha )$ we know that there is one minimum. We'll find it by setting the derivative of $D_f(\alpha )$ equal to 0. We have $$D_f(\alpha )={180}^{\circ }+2\alpha -4\beta ,$$ where $$\beta=\beta(\alpha)=arcsin{\frac{\sin{\alpha}}{n_{f,w}}.$$Tomakethingslooksimplerw{e}^{\prime }llwrite\$n\$for\$n_{f,w}.\$Now$$D_f^\prime(\alpha)=2-4\beta^\prime(\alpha).$$Thederivativeofthefunction\$f(x)=\arcsin (x)\$is$$f^\prime(x)=\frac{1}{\sqrt{1-x^2}}.$$Usingthechainrulegives$$\beta^\prime(\alpha)=\frac{\cos{\alpha}}{n\sqrt{1-\frac{\sin^2{\alpha}}{n^2}}}=\frac{\cos{\alpha}}{\sqrt{n^2-\sin^2{\alpha}}}.$$Thus$$D_f^\prime(\alpha)=2-\frac{4\cos{\alpha}}{\sqrt{n^2-\sin^2{\alpha}}}.$$Setting\$D_f(\alpha )=0\$gives$$2-\frac{4\cos{\alpha}}{\sqrt{n^2-\sin^2{\alpha}}}=0.$$Notingthat\${\sin }^2\alpha =1-{\cos }^2\alpha \$andrearranginggives$$\cos^2{\alpha}=\frac{n^2-1}{3}.$Thus,$D_f^\prime(\alpha)=0$for$${\alpha }_m=\arccos \sqrt{\frac{n^2-1}{3}}.$$

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