Question

pllzz answer it!!!!!! A quadratic equation whose roots are sin ^2 18 ,cos^2 36 is

Answer 1

$\huge\underline\bold\orange{Answer}$

$\rule{200}{20}$

$\huge\underline\bold\blue{Given}$

$Roots\:of\:a\: quadratic\: equation\:are\:sin^2\:18°\:,cos^2\:36°\:are$

$\huge\underline\bold\red{To \: find}$

$The\: quadratic\: equation$

$\huge\underline\bold\green{Solution}$

$\color{red}{Solve\:the\:Sum\:of\: the\:roots}$

$=\: sin^2\:18°\:+\:cos^2\:36°$

$=\:\frac{\sqrt[5]\:-\:1}{4}^2\:+\:\frac{\sqrt[5]\:+\:1}{4}^2$

$=\:\frac{1}{16}\:×\:\sqrt{[5]\:-\:1}^2(\sqrt{[5]\:+\:1}^2)$

$=\:\frac{1}{16}\:×\:[2(5\:+\:1)$

$=\:\frac{3}{4}$

$\color{purple}{Product\:of\:roots}$

$=\:sin^2\:18\:.\:cos^2\:36°$

$=\:\frac{\sqrt[5]\:-\:1}{4}^2\:.\:\frac{\sqrt[5]\:+\:1}{4}^2$

$=\:\frac{5-4}{4.4}^2\:=\:\frac{1}{16}$

$=\: Therefore\: quadratic\: equation\:is$

$=\:x^2\:-\:[sin^2\:18°\:+\:cos^2\:36°]x\:+\:[sin^2\:18°\:cos^2\:36°]$

$=>\:x^2\:+\:[\frac{3}{4}]x\:+\:\frac{1}{16}$

$=>\:16x^2\:-\:12x\:+\:1\:=\:0$

$\rule{200}{20}$

$\huge\underline\bold\pink{Thanks}$

Answer 2

Answer:

pls mark me as brainliest & thank me