Question

$\begin{array}{cc}\text{Column 1}& \text{Column 2}\\ \text{1.sin}{18}^{\circ }& \text{p.}\frac{-\sqrt{5}-1}{4}\\ \text{2.cos}{18}^{\circ }& \text{q.}\frac{-1+\sqrt{5}}{4}\\ \text{3.cos}{9}^{\circ }\text{- sin}{9}^{\circ }& \text{r.}\frac{\sqrt{10+2\sqrt{5}}}{4}\\ & \text{s.}\frac{10+2\sqrt{5}}{4}\\ & \text{t.}\frac{\sqrt{5-\sqrt{5}}}{2}\\ & \text{u.}\frac{\sqrt{5-2\sqrt{5}}}{2}\end{array}$

• A. 1-q 2-r 3-t
• B. 1-r 2-t 3-q
• C. 1-p 2-r 3-4
• D. 1-q 2-t 3-u

Answer 1

The correct option is A

1-q 2-r 3-t

1.$sin{18}^{\circ }$

We can't write sin18 as compound angle (A $\pm$ B) from where A & B are standard angles (0,30,45,60,90)

We observe that,

18 $\times ={90}^{\circ }$

Let $\theta ={18}^{\circ }$

$5\theta ={90}^{\circ }$

$2\theta +3\theta ={90}^{\circ }$

$2\theta ={90}^{\circ }-+3\theta$

$sin2\theta =sin(90-3\theta )$

$2sin\theta .cos\theta =cos3\theta =4co{s}^3\theta -3cos\theta$

$2sin\theta .cos\theta =cos\theta (4co{s}^2\theta -3)$

$(2sin\theta -4co{s}^2\theta +3)cos\theta =0$

{$Ifcos\theta =0\theta ={90}^{\circ }Butwehavetaken\theta ={18}^{\circ }cos\theta cannotbeequaltozero$}

So,$cos\theta \ne 0$

$2sin\theta -4co{s}^2\theta +3=0$

$2sin\theta -4(1-si{n}^2\theta )-3=0$

$2sin\theta -4+4si{n}^2\theta +3=0$

$4si{n}^2\theta +2sin\theta -1=0$

$sin\theta =\frac{-2\pm \sqrt{(2{)}^2-(4\times 4\times (-1))}}{2\times 4}$

= $\frac{-2\pm \sqrt{4+16}}{8}$

= $\frac{-2\pm \sqrt{20}}{8}$ = $\frac{-2\pm 2\sqrt{5}}{8}$ = $\frac{-1\pm \sqrt{5}}{4}$

$\theta ={18}^{\circ }sin\theta cannotbenegative$

$sin{18}^{\circ }=\frac{-1+\sqrt{5}}{4}$

2. In $△ABC$

$A{B}^2=16-(-1+\sqrt{5}{)}^2$

= $16-(1+5-2\sqrt{5})$

= $16-6+2\sqrt{5}$

= $10+2\sqrt{5}$

AB = $\sqrt{10+2\sqrt{5}}$

$cos{18}^{\circ }=\frac{AB}{AC}=\frac{\sqrt{10+2\sqrt{5}}}{4}$

3. $cos{9}^{\circ }-sin{9}^{\circ }$

Letx = $cos{9}^{\circ }-sin{9}^{\circ }$

On squaring both sides, we will get the terms of $cos{9}^{\circ }-sin{9}^{\circ }and2cos{9}^{\circ }sin{9}^{\circ }$. These terms can be written as 1 & $sin{18}^{\circ }$ respectively.

$x^2=co{s}^2{9}^{\circ }+si{n}^2{9}^{\circ }-2cos{9}^{\circ }sin{9}^{\circ }$

= $1-sin{18}^{\circ }$ {$sin{18}^{\circ }=\frac{\sqrt{5}-1}{4}$}

= $1-\frac{\sqrt{5}-1}{4}$

= $\frac{4-\sqrt{5}+1}{4}=\frac{5+\sqrt{5}}{4}$

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

$cos{9}^{\circ }-sin{9}^{\circ }=\frac{\sqrt{5-\sqrt{5}}}{2}$