[tex] a + b = 2\pi\implies \: a = 2\pi - b \\ \\ \sin(a)\cdot\sin(b) \leq0 \\ \sin(2\pi - b)\cdot\sin(b) \leq0 \\ \sin( - b + 2\pi)\cdot\sin(b) \leq0 \\ \sin( - b)\cdot\sin(b) \leq0 \\ - \sin(b) \cdot\sin(b) \leq0 \\ - \sin(b) {}^{2} \leq0 \\ \sin(b) {}^{2} \: \: \text{este \: mereu \: pozitivă \: sau \: 0} \\ \implies - \sin(b) {}^{2} \leq0 \: \: \text{este \: adevarata} \\ \implies \: \boxed{b\in\mathbb{R}} [/tex]
[tex] \text{la \ \: partea \: cu \: sin(} - b + 2\pi) \\ \text{am \: folosit \: formula : \: } \\ \boxed{\sin(t\pm2k\pi) = \sin(t) \: , \: k\in\mathbb{Z}}[/tex]
Show life that you have a thousand reasons to smile
© Copyright 2024 DOKU.TIPS - All rights reserved.
[tex] a + b = 2\pi\implies \: a = 2\pi - b \\ \\ \sin(a)\cdot\sin(b) \leq0 \\ \sin(2\pi - b)\cdot\sin(b) \leq0 \\ \sin( - b + 2\pi)\cdot\sin(b) \leq0 \\ \sin( - b)\cdot\sin(b) \leq0 \\ - \sin(b) \cdot\sin(b) \leq0 \\ - \sin(b) {}^{2} \leq0 \\ \sin(b) {}^{2} \: \: \text{este \: mereu \: pozitivă \: sau \: 0} \\ \implies - \sin(b) {}^{2} \leq0 \: \: \text{este \: adevarata} \\ \implies \: \boxed{b\in\mathbb{R}} [/tex]
[tex] \text{la \ \: partea \: cu \: sin(} - b + 2\pi) \\ \text{am \: folosit \: formula : \: } \\ \boxed{\sin(t\pm2k\pi) = \sin(t) \: , \: k\in\mathbb{Z}}[/tex]