Q55 / 160 · TS EAMCET 2023 (14 May Shift 1) · Mathematics

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

  • A
    $16 x^2-12 x-1=0$
  • B
    $16 x^2-12 x+4=0$
  • Correct answer
    C
    $16 x^2-12 x+1=0$
  • D
    $16 x^2+12 x+1=0$

Solution

$\sin ^2 18^{\circ}=\left(\frac{\sqrt{5}-1}{4}\right)^2=\frac{6-2 \sqrt{5}}{16}=\frac{3-\sqrt{5}}{8}$ $$ \cos ^2 36^{\circ}=\left(\frac{\sqrt{5}+1}{4}\right)^2=\frac{3+\sqrt{5}}{8} $$ $\therefore$ Required equation is $$ \begin{aligned} & x^2-\left(\frac{3-\sqrt{5}}{8}+\frac{3+\sqrt{5}}{8}\right) x+\left(\frac{3-\sqrt{5}}{8} \times \frac{3+\sqrt{5}}{8}\right)=0 \\ & x^2-\frac{3}{4} x+\frac{1}{16}=0 \Rightarrow 16 x^2-12 x+1=0 \end{aligned} $$

Hence, the correct answer is option: 'C'.

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