Question

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

• A. $16x^2-12x+1=0$
• B. $16x^2+12x+1=0$
• C. $16x^2-20x+5=0$
• D. $16x^2+20x+5=0$

Answer 1

The correct option is C $16x^2-20x+5=0$

We have,

$\begin{array}{c}{\cos }^2{18}^0+{\sin }^2{36}^0=\frac{1}{2}\left[1+\cos {36}^0+1-\cos {72}^0\right]\\ =\frac{1}{2}\left[2+\frac{\sqrt{5+1}}{4}-\frac{\sqrt{5-1}}{4}\right]\\ =\frac{1}{2}\left[2+\frac{1}{2}\right]=\frac{5}{4}\end{array}$

Out of given options only Option $C$ has sum of its roots as $5/4$.

Product of roots ${\cos }^2{18}^0+{\sin }^2{36}^0=\frac{1}{4}{\left(\sin {54}^0+\sin {18}^0\right)}^2$

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

$=\frac{1}{4}{\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{5}{16}$

Hence,

Option $C$ is correct answer.