Question

lf ${\cos }^2\left(\frac{\pi }{8}\right)$ is a root of the equation $x^2+ax+b=0$, where $a,b$ are rational numbers, then ordered pair $(a,b)$, is:

• A. $(1,-\frac{1}{8})$
• B. $(-1,-\frac{1}{8})$
• C. $(-1,\frac{1}{8})$
• D. $(1,\frac{1}{8})$

Answer 1

Answer: (C)

$(-1,\frac{1}{8})$

$2{\cos }^2\left(\frac{\pi }{8}\right)-1={\cos }^2\left(\frac{\pi }{4}\right)$
$\because \cos 2\theta =2{\cos }^2\theta -1.$
$\Rightarrow {\cos }^2\frac{\pi }{8}=\frac{{\cos }^2\left(\frac{\pi }{4}\right)+1}{2}$
$=\frac{\frac{1}{\sqrt{2}}+1}{2}=\frac{\sqrt{2}+1}{2\sqrt{2}}$
Since $a,b$ are rational, other root is its rational conjugate , i.e $\frac{\sqrt{2}-1}{2\sqrt{2}}$
$\therefore a=-(\frac{\sqrt{2}+1}{2\sqrt{2}}+\frac{\sqrt{2}-1}{2\sqrt{2}})=-1$
$\therefore b=\left(\frac{\sqrt{2}+1}{2\sqrt{2}}\right)\left(\frac{\sqrt{2}-1}{2\sqrt{2}}\right)=\frac{1}{8}$