[tex]b)\\\tan x=\dfrac{\sin x}{c\os x}\\\\\tan^2x=\dfrac{\sin^2x}{\cos^2x}=\dfrac{2\sin^2x}{2\cos^2x}=\dfrac{\sin^2x+\sin^2x}{\cos^2x+\cos^2x}\\\\=\dfrac{\sin^2x+\sin^2x+\cos^2x-\cos^2x}{\cos^2x+\cos^2x+\sin^2x-\sin^2x}=\dfrac{(\sin^2x+\cos^2x)-(\cos^2x-\sin^2x)}{(\cos^2x+\sin^2x)+(\cos^2x-\sin^2x)}\\\\=\dfrac{1-\cos2x}{1+\cos2x}\\\\\text{Used}:\\\\\sin^2x+\cos^2x=1\\\\\cos^2x-\sin^2x=\cos2x[/tex]
[tex]\tan^2\left(\dfrac{7\pi}{8}\right)=\dfrac{1-\cos\left(2\cdot\dfrac{7\pi}{8}\right)}{1+\cos\left(2\cdot\dfrac{7\pi}{8}\right)}=\dfrac{1-\cos\left(\dfrac{7\pi}{4}\right)}{1+\cos\left(\dfrac{7\pi}{4}\right)}=(*)\\\\\cos\left(\dfrac{7\pi}{4}\right)=\cos\left(2\pi-\dfrac{\pi}{4}\right)=\cos\left(-\dfrac{\pi}{4}\right)=\cos\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt2}{2}\\\\(*)=\dfrac{1-\frac{\sqrt2}{2}}{1+\frac{\sqrt2}{2}}=\left(\dfrac{2}{2}-\dfrac{\sqrt2}{2}\right):\left(\dfrac{2}{2}+\dfrac{\sqrt2}{2}\right)[/tex]
[tex]=\dfrac{2-\sqrt2}{2}\div\dfrac{2+\sqrt2}{2}=\dfrac{2-\sqrt2}{2}\cdot\dfrac{2}{2+\sqrt2}=\dfrac{2-\sqrt2}{2+\sqrt2}\\\\\text{Use}\ (a+b)(a-b)=a^2-b^2\\\\=\dfrac{2-\sqrt2}{2+\sqrt2}\cdot\dfrac{2-\sqrt2}{2-\sqrt2}=\dfrac{(2-\sqrt2)^2}{2^2-(\sqrt2)^2}\\\\\text{Use}\ (a-b)^2=a^2-2ab+b^2\\\\=\dfrac{2^2-2(2)(\sqrt2)+(\sqrt2)^2}{4-2}=\dfrac{4-4\sqrt2+2}{2}=2-2\sqrt2+1\\\\=(\sqrt2)^2-2(\sqrt2)(1)+1^2=1^2-2(1)(\sqrt2)+(\sqrt2)^2\\\\\text{Use}\ (a-b)^2=a^2-2ab+b^2\\\\=(1-\sqrt2)^2\\\\\text{Therefore}:[/tex]
[tex]\tan^2\left(\dfrac{7\pi}{8}\right)=(1-\sqrt2)^2\to\boxed{\tan\left(\dfrac{7\pi}{8}\right)=1-\sqrt2}[/tex]
[tex]c)\\\tan\left(\dfrac{\pi}{8}\right)=\tan\left(\pi-\dfrac{7\pi}{8}\right)=\tan\left(-\dfrac{7\pi}{8}\right)=-\tan\left(\dfrac{7\pi}{8}\right)\\\\=-(1-\sqrt2)=\sqrt2-1\\\\y=\sin(2x-1)+a\tan\dfrac{\pi}{8}\\\\\text{We know}\ -1\leq\sin(2x-1)\leq1.\\\\y\geq0\ \text{therefore}\ a\tan\dfrac{\pi}{8}\geq1\\\\\text{We have to move the graph at least one unit up}\\\\a(\sqrt2-1)\geq1\qquad\text{divide both sides by}\ (\sqrt2-1)>0\\\\a\geq\dfrac{1}{\sqrt2-1}\cdot\dfrac{\sqrt2+1}{\sqrt2+1}\\\\a\geq\dfrac{\sqrt2+1}{(\sqrt2)^2-1^2}[/tex]
[tex]a\geq\dfrac{\sqrt2+1}{2-1}\\\\a\geq\dfrac{\sqrt2+1}{1}\\\\a\geq\sqrt2+1\\\\Answer:\ \boxed{a=\sqrt2+1}[/tex]
