Question

The quadratic equation whose roots are ${\sin }^{2}18^0$ and ${\cos }^{2}36^0$ is

• A. $16x^2-12x+1=0$
• B. $16x^2+12x+1=0$
• C. $16x^2-12x-1=0$
• D. $16x^2-10x+1=0$

Answer 1

Answer: (A)

Sum of roots $={\sin }^{2}18+{\cos }^{2}36$
$={\left(\frac{\sqrt{5}-1}{4}\right)}^2+{\left(\frac{\sqrt{5}+1}{4}\right)}^2$
$=\frac{5+1-2\sqrt{5}+5+1+2\sqrt{5}}{16}=\frac{12}{16}=\frac{3}{4}$
Product $={\sin }^{2}18\cdot {\cos }^{2}36$
$={\left(\frac{\sqrt{5}-1}{4}\right)}^2\cdot {\left(\frac{\sqrt{5}+1}{4}\right)}^2$
$={\left(\frac{5-1}{16}\right)}^2={\left(\frac{4}{16}\right)}^2=\frac{1}{16}$
$=$ equations is $x^2-\frac{3x}{4}+\frac{1}{16}=0$
$\Rightarrow 16x^2-12x+1=0$