Question

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

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

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 $x^{2}-\frac{3 x}{4}+\frac{1}{16}=0$
$\Rightarrow 16 x^{2}-12 x+1=0$