Prove that. \(sin^2 \frac{π}{6}+cos^2 \frac{π}{3}–tan^2 \frac{π}{4}=-\frac{1}{2}\)
Find \(sin \frac{x}{2},\,cos \frac{x}{2} \,and \,tan\,\frac{ x}{2}\, for \,sin\,\,x=\frac{1}{4}\) in quadrant II.