Question

The value of $\cos 9^{\circ }-\sin 9^{\circ }$ is

• A. $-\frac{\sqrt{5-\sqrt{5}}}{2}$
• B. $\frac{5+\sqrt{5}}{4}$
• C. $\frac{1}{2}\sqrt{5-\sqrt{5}}$
• D. none of these.

Answer 1

Answer: (C)

$\frac{1}{2}\sqrt{5-\sqrt{5}}$

Consider, $\sin {72}^{}=2\sin {36}^{}\cos {36}^{}$

$\Rightarrow \sin ({90}^{}-{18}^{})=4\sin {18}^{}\cos {18}^{}(1-2{\sin }^2{18}^{})$

$\Rightarrow \cos {18}^{}=4\sin {18}^{}\cos {18}^{}(1-2{\sin }^2{18}^{})$

$\Rightarrow 8{\sin }^3{18}^{}-4\sin {18}^{}+1=0$

$\Rightarrow (2\sin {18}^{}-1)(4{\sin }^2{18}^{}+2\sin {18}^{}-1)=0\because \sin {18}^{}\ne \frac{1}{2}$

$\therefore 4{\sin }^2{18}^{}+2\sin {18}^{}-1=0$

$\Rightarrow \sin {18}^{}=\frac{-1\pm \sqrt{5}}{4}$
$a=\cos 9^{}-\sin 9^{}$

$\Rightarrow a^2={\cos }^{2}9+{\sin }^2{9}^{}-2\cos 9^{}\sin 9^{}$

$\Rightarrow a^2={\cos }^{2}9+{\sin }^2{9}^{}-\sin {18}^{}$

$\Rightarrow a^2=1-\sin {18}^{}$

$\Rightarrow a^2=1-\left(\frac{-1+\sqrt{5}}{4}\right)$

$\Rightarrow a^2=\frac{5-\sqrt{5}}{4}$

$\therefore a=\frac{1}{2}\sqrt{5-\sqrt{5}}$