Question

The quadratic equation whose roots are $si{n}^2{18}^{\circ }$ and $co{s}^2{36}^{\circ }$ is

• A. 16x² - 12x + 1 = 0
• B. 16x² + 12x + 1 = 0
• C. 16x² - 12x - 1 = 0
• D. 16x² + 10x + 1 = 0

Answer 1

The correct option is A

16x² - 12x + 1 = 0

Since $si{n}^2{18}^{\circ }$ and $co{s}^2{36}^{\circ }$ are the roots of quadratic equation

$\therefore$ sum of roots = $si{n}^2{18}^{\circ }$ + $co{s}^2{36}^{\circ }$

= ${\left(\frac{\sqrt{5}-1}{4}\right)}^2$ + ${\left(\frac{\sqrt{5}+1}{4}\right)}^2$ = $\frac{3}{4}$

And products of roots = $si{n}^2{18}^{\circ }$.$co{s}^2{36}^{\circ }$

= ${\left(\frac{\sqrt{5}-1}{4}\right)}^2$.${\left(\frac{\sqrt{5}+1}{4}\right)}^2$ = $\frac{1}{16}$

$\therefore$ Required equation whose roots are $si{n}^2{18}^{\circ }$ and $co{s}^2{36}^{\circ }$ is

$x^2-\frac{3}{4}x+\frac{1}{16}x=0\Rightarrow 16x^2-12x+1=0$