Question

I need help with part c). Given that tan(pi/8)= sqrt(2)-1. Please provide a detailed answer.

Answer 1

`b)\\\tan x=(\sin x)/(c\os x)\\\\\tan^2x=(\sin^2x)/(\cos^2x)=(2\sin^2x)/(2\cos^2x)=(\sin^2x+\sin^2x)/(\cos^2x+\cos^2x)\\\\=(\sin^2x+\sin^2x+\cos^2x-\cos^2x)/(\cos^2x+\cos^2x+\sin^2x-\sin^2x)=((\sin^2x+\cos^2x)-(\cos^2x-\sin^2x))/((\cos^2x+\sin^2x)+(\cos^2x-\sin^2x))\\\\=(1-\cos2x)/(1+\cos2x)\\\\\text{Used}:\\\\\sin^2x+\cos^2x=1\\\\\cos^2x-\sin^2x=\cos2x`

`\tan^2\left((7\pi)/(8)\right)=(1-\cos\left(2\cdot(7\pi)/(8)\right))/(1+\cos\left(2\cdot(7\pi)/(8)\right))=(1-\cos\left((7\pi)/(4)\right))/(1+\cos\left((7\pi)/(4)\right))=(*)\\\\\cos\left((7\pi)/(4)\right)=\cos\left(2\pi-(\pi)/(4)\right)=\cos\left(-(\pi)/(4)\right)=\cos\left((\pi)/(4)\right)=(\sqrt2)/(2)\\\\(*)=(1-(\sqrt2)/(2))/(1+(\sqrt2)/(2))=\left((2)/(2)-(\sqrt2)/(2)\right):\left((2)/(2)+(\sqrt2)/(2)\right)`

`=(2-\sqrt2)/(2)/(2+\sqrt2)/(2)=(2-\sqrt2)/(2)\cdot(2)/(2+\sqrt2)=(2-\sqrt2)/(2+\sqrt2)\\\\\text{Use}\ (a+b)(a-b)=a^2-b^2\\\\=(2-\sqrt2)/(2+\sqrt2)\cdot(2-\sqrt2)/(2-\sqrt2)=((2-\sqrt2)^2)/(2^2-(\sqrt2)^2)\\\\\text{Use}\ (a-b)^2=a^2-2ab+b^2\\\\=(2^2-2(2)(\sqrt2)+(\sqrt2)^2)/(4-2)=(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}:`

`\tan^2\left((7\pi)/(8)\right)=(1-\sqrt2)^2\to\boxed{\tan\left((7\pi)/(8)\right)=1-\sqrt2}`

`c)\\\tan\left((\pi)/(8)\right)=\tan\left(\pi-(7\pi)/(8)\right)=\tan\left(-(7\pi)/(8)\right)=-\tan\left((7\pi)/(8)\right)\\\\=-(1-\sqrt2)=\sqrt2-1\\\\y=\sin(2x-1)+a\tan(\pi)/(8)\\\\\text{We know}\ -1\leq\sin(2x-1)\leq1.\\\\y\geq0\ \text{therefore}\ a\tan(\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(1)/(\sqrt2-1)\cdot(\sqrt2+1)/(\sqrt2+1)\\\\a\geq(\sqrt2+1)/((\sqrt2)^2-1^2)`

`a\geq(\sqrt2+1)/(2-1)\\\\a\geq(\sqrt2+1)/(1)\\\\a\geq\sqrt2+1\\\\Answer:\ \boxed{a=\sqrt2+1}`