NIMCET 2018 — Mathematics PYQ

Concept:

• Trigonometric Values:
  $\sin 18^{\circ }=\frac{\sqrt{5}-1}{4}$  
  $\cos 36^{\circ }=\frac{\sqrt{5}+1}{4}$  

• The quadratic equation whose roots are $\alpha$ and $\beta$ is given by:  
  $(x-\alpha )(x-\beta )=0$  
  $\Rightarrow x^2-(\alpha +\beta )x+\alpha \beta =0$

Calculation

Let's say that the roots of the quadratic equation are $\alpha ={\sin }^{2}18^{\circ }$ and $\beta ={\cos }^{2}36^{\circ }$.

$\alpha ={\sin }^{2}18^{\circ }={\left(\frac{\sqrt{5}-1}{4}\right)}^2=\frac{6-2\sqrt{5}}{16}=\frac{3-\sqrt{5}}{8}$

$\beta ={\cos }^{2}36^{\circ }={\left(\frac{\sqrt{5}+1}{4}\right)}^2=\frac{6+2\sqrt{5}}{16}=\frac{3+\sqrt{5}}{8}$

Now,  
$\alpha +\beta =\frac{3-\sqrt{5}}{8}+\frac{3+\sqrt{5}}{8}=\frac{3}{4}$

And,  
$\alpha \beta =\left(\frac{3-\sqrt{5}}{8}\right)\left(\frac{3+\sqrt{5}}{8}\right)=\frac{9-5}{64}=\frac{1}{16}$

∴ The required equation is:  
$x^2-(\alpha +\beta )x+\alpha \beta =0$  

$\Rightarrow x^2-\left(\frac{3}{4}\right)x+\frac{1}{16}=0$

On multiplying by 16, we get:  
$\Rightarrow 16x^2-12x+1=0$