Question

Sa se demonstreze ca

1-sin^2 x•cos x+cos x-sin^2 x>=0

Answer 1

$1-\sin^2 x\cos x+\cos x-\sin^2 x=\\ \\ =1-\cos x(\sin^2 x-1)-\sin^2 x \\ \\ =1-\cos x\cdot (-\cos^2 x)-(1-\cos^2 x) \\ \\ =1+\cos^3 x+\cos^2 x-1\\ \\ =\cos^3 x+\cos^2 x \\ \\ = \cos^2 x(\cos x+1)\\ \\ \text{Cum }-1\leq \cos x\leq 1\Leftrightarrow 0\leq \cos x\leq 2\\ \text{Si }\cos^2 x\geq 0. \\ \\ \Rightarrow \cos^2 x(\cos x+1)\geq 0\quad (q.e.d.)$