Question

If $4{\cos }^2{x}^0-1=0$ and $0\le x^0\le {90}^0$, find ${\sin }^2{x}^0+{\cos }^2{x}^0$.

Answer 1

$4co{s}^2{x}^{\circ }-1=0$
$co{s}^2{x}^{\circ }=\frac{1}{4}$
$cos{x}^{\circ }=\frac{1}{2}$
$cos{x}^{\circ }=cos{60}^{\circ }$
$x={60}^{\circ }$
Now, $si{n}^2{x}^{\circ }+co{s}^2{x}^{\circ }$
= $si{n}^2{60}^{\circ }+co{s}^2{60}^{\circ }$
= $(\frac{\sqrt{3}}{2}{)}^2+(\frac{1}{2}{)}^2$
= $\frac{3}{4}+\frac{1}{4}$
= $\frac{4}{4}$
= $1$