Aspire Faculty ID #3750 · Topic: NIMCET 2018 · Just now
NIMCET 2018

The quadratic equation whose roots are $\sin^218^o$ and $\cos^236^o$  is

Solution

Roots: 
$\alpha = \sin^2 18^\circ = \left(\frac{\sqrt{5} - 1}{4}\right)^2 = \frac{3 - \sqrt{5}}{8}$ 
$\beta = \cos^2 36^\circ = \left(\frac{\sqrt{5} + 1}{4}\right)^2 = \frac{3 + \sqrt{5}}{8}$ 

Sum of roots ($S$): 
$S = \alpha + \beta = \frac{(3 - \sqrt{5}) + (3 + \sqrt{5})}{8} = \frac{6}{8} = \frac{3}{4}$ 

Product of roots ($P$): $P = \alpha\beta = \left(\frac{(\sqrt{5} - 1)(\sqrt{5} + 1)}{16}\right)^2 = \left(\frac{4}{16}\right)^2 = \frac{1}{16}$ 

Equation: $x^2 - Sx + P = 0$ $x^2 - \frac{3}{4}x + \frac{1}{16} = 0 \implies 16x^2 - 12x + 1 = 0$

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