от ammornil » 24 Апр 2024, 13:13
Гост написа:А= sin10a+ sin2a+ sin8a+ sin4a
Благодаря предварително!!!
[tex]\sin{\beta}+\sin{\gamma} = 2\cdot{}\sin{\frac{\beta+\gamma}{2}}\cdot{}\cos{\frac{\beta-\gamma}{2}} \\ \quad \\ \quad \sin{10\alpha}+\sin{2\alpha}=2\cdot{}\sin{\frac{10\alpha+2\alpha}{2}}\cdot{}\cos{\frac{10\alpha-2\alpha}{2}}=2\cdot{}\sin{6\alpha}\cdot{}\cos{8\alpha} \\ \quad \sin{8\alpha}+\sin{4\alpha}=2\cdot{}\sin{\frac{8\alpha+4\alpha}{2}}\cdot{}\cos{\frac{8\alpha-4\alpha}{2}}=2\cdot{}\sin{6\alpha}\cdot{}\cos{2\alpha} \\ \quad \\ A=2\cdot{}\sin{6\alpha}\cdot{}\cos{8\alpha}+2\cdot{}\sin{6\alpha}\cdot{}\cos{2\alpha}=2\cdot{}\sin{6\alpha}\cdot{}(\cos{8\alpha}+\cos{2\alpha}) \\ \quad \\ \quad \cos{8\alpha}+\cos{2\alpha}=2\cdot{}\cos{\frac{8\alpha+2\alpha}{2}}\cdot{} \cos{\frac{8\alpha-2\alpha}{2}}=2\cdot{}\cos{5\alpha}\cdot{}\cos{3\alpha} \\ \quad \\ A=2\cdot{}\sin{6\alpha}\cdot{}2\cdot{}\cos{5\alpha}\cdot{}\cos{3\alpha}=4\cdot{}\sin{6\alpha}\cdot{}\cos{5\alpha}\cdot{}\cos{3\alpha}[/tex]
[tex]\color{lightseagreen}\text{''Който никога не е правил грешка, никога не е опитвал нещо ново.''} \\
\hspace{21em}\text{(Алберт Айнщайн)}[/tex]