Question

If $a$ and $b$ are the roots of the equation $x^2−4x+1=0(a>b)$ then the value of
$f\left(Î\pm ,β\right)=\frac{{β}^3}{2}cose{c}^2\left(\frac{1}{2}ta{n}^{−1}\frac{β}{Î\pm }\right)+\frac{{Î\pm }^3}{2}se{c}^2\left(\frac{1}{2}ta{n}^{−1}\frac{Î\pm }{β}\right)$ is

Answer 1

Answer: 56

$f\left(Î\pm ,β\right)=\frac{{β}^3}{2}cose{c}^2\left(\frac{1}{2}ta{n}^{−1}\frac{β}{Î\pm }\right)+\frac{{Î\pm }^3}{2}se{c}^2\left(\frac{1}{2}ta{n}^{−1}\frac{Î\pm }{β}\right)$

Let $ta{n}^{−1}\left(\frac{β}{Î\pm }\right)=θ$ and $ta{n}^{−1}\left(\frac{Î\pm }{β}\right)=Ï•$
$f\left(Î\pm ,β\right)=\frac{{β}^3}{2si{n}^2\frac{θ}{2}}+\frac{{Î\pm }^3}{2co{s}^2\frac{Ï•}{2}}=\frac{{β}^3}{1−cosθ}+\frac{{Î\pm }^3}{1+cosÏ•}$
$=\frac{{β}^3}{1−\frac{Î\pm }{\sqrt{{Î\pm }^2+{β}^2}}}+\frac{{Î\pm }^3}{1+\frac{β}{\sqrt{{Î\pm }^2+{β}^2}}}$
$=\sqrt{{Î\pm }^2+{β}^2}\left[\frac{{β}^3}{\sqrt{{Î\pm }^2+{β}^2}−Î\pm }+\frac{{Î\pm }^3}{\sqrt{{Î\pm }^2+{β}^2}+β}\right]$
$=\sqrt{{Î\pm }^2+{β}^2}\left[\frac{{β}^3\left(\sqrt{{Î\pm }^2+{β}^2}+Î\pm \right)}{{β}^2}+\frac{{Î\pm }^3\left(\sqrt{{Î\pm }^2+{β}^2−β}\right)}{{Î\pm }^2}\right]$
$=\sqrt{{Î\pm }^2+{β}^2}\left[β\sqrt{{Î\pm }^2+{β}^2}+Î\pm \sqrt{{Î\pm }^2+{β}^2}\right]$
$f\left(Î\pm ,β\right)=\left({Î\pm }^2+{β}^2\right)\left(Î\pm +β\right)$
Now, $Î\pm +β=4$ and $Î\pm β=1$
$f\left(Î\pm ,β\right)=\left({\left(Î\pm +β\right)}^2−2Î\pm β)\right)\left(Î\pm +β\right)$
$=\left(16−2\right)\left(4\right)=56$