Question

Given that ${\sin }^{10}x+{\cos }^{10}x=\frac{29}{64},$ if ${\sin }^{12}x+{\cos }^{12}x=\frac{a}{b}$, $a$ and $b$ are coprime, then the value of $b-a$ is

• A. $155$
• B. $156$
• C. $157$
• D. $158$

Answer 1

The correct option is C $157$

${\sin }^{10}x+{\cos }^{10}x=\frac{29}{64}$
Let ${\sin }^{2}x=\frac{1}{2}+t,|t|\le \frac{1}{2}$
$\Rightarrow {\cos }^{2}x=\frac{1}{2}-t$

$\Rightarrow {(\frac{1}{2}+t)}^5+{(\frac{1}{2}-t)}^5=\frac{29}{64}\Rightarrow 2(\frac{1}{32}+\frac{10}{8}{t}^2+\frac{5}{2}{t}^4)=\frac{29}{64}\Rightarrow 4+160t^2+320t^4=29\Rightarrow 64t^4+32t^2-5=0\Rightarrow t^2=\frac{1}{8}$

${\sin }^{12}x+{\cos }^{12}x={(\frac{1}{2}+t)}^6+{(\frac{1}{2}-t)}^6=\frac{1}{32}[1+60t^2+240t^4+64t^6]=\frac{99}{256}$