Здравейте, ще може ли някой да реши тази задача, б подточка

[Гост]

благодаря предварително

[ammornil]

[tex]\cos{10^{\circ}}\cdot{\cos{30^{\circ}}}\cdot{\cos{50^{\circ}}}\cdot{\cos{70^{\circ}}}=?[/tex]

$$ \begin{matrix} f(x) & 0^{\circ} & 30^{\circ} & 45^{\circ} & 60^{\circ} & 90^{\circ} \\ \cos(x) & 1 & \frac{\sqrt{3}}{2} & \frac{\sqrt{2}}{2} & \frac{1}{2} & 0 \end{matrix} $$

$$ \cos{\alpha}\cdot{\cos{\beta}} = \frac{1}{2}[\cos{(\alpha+\beta)}+\cos{(\alpha-\beta)}]; \hspace{2em} \cos{(180^{\circ}-\alpha)}=-\cos{\alpha}$$

[tex]\cos{50^{\circ}}\cdot{\cos{10^{\circ}}} = \frac{1}{2}[\cos{(50^{\circ}+10^{\circ})}+\cos{(50^{\circ}-10^{\circ})}]= \frac{1}{2}[\cos{60^{\circ}}+\cos{40^{\circ}}][/tex]

[tex]\cos{10^{\circ}}\cdot{\cos{50^{\circ}}}\cdot{\cos{70^{\circ}}}=\frac{1}{2}(\cos{60^{\circ}}+\cos{40^{\circ}})\cdot{\cos{70^{\circ}}}=\frac{1}{2}[\cos{60^{\circ}}\cdot{\cos{70^{\circ}}+\cos{40^{\circ}}\cdot{\cos{70^{\circ}}}}][/tex]

[tex]\hspace{3em}\cos{40^{\circ}}\cdot{\cos{70^{\circ}}}=\frac{1}{2}[\cos{(70^{\circ}+40^{\circ})}+\cos{(70^{\circ}-40^{\circ})}]= \frac{1}{2}[\cos{110^{\circ}}+\cos{30^{\circ}}][/tex]

[tex]\hspace{5em}\cos{110^{\circ}}=\cos{(180^{\circ}-70^{\circ})}=-\cos{70^{\circ}} \Rightarrow \frac{1}{2}(\cos{110^{\circ}}+\cos{30^{\circ}})=\frac{1}{2}(\cos{30^{\circ}}-\cos{70^{\circ}}) \Rightarrow[/tex]

[tex]\Rightarrow \cos{10^{\circ}}\cdot{\cos{50^{\circ}}}\cdot{\cos{70^{\circ}}}=\frac{1}{2}[\cos{60^{\circ}}\cdot{\cos{70^{\circ}}+\frac{1}{2}(\cos{30^{\circ}}-\cos{70^{\circ}})}]=\frac{1}{2}\cdot{\frac{1}{2}}\cdot{\cos{70^{\circ}}}+\frac{1}{4}(\frac{\sqrt{3}}{2}-\cos{70^{\circ}})=\frac{1}{4}\cos{70^{\circ}}+\frac{\sqrt{3}}{8}-\frac{1}{4}\cos{70^{\circ}}= \frac{\sqrt{3}}{8}[/tex]

[tex]\Rightarrow \cos{30^{\circ}}\cdot{\cos{10^{\circ}}}\cdot{\cos{50^{\circ}}}\cdot{\cos{70^{\circ}}}=\frac{\sqrt{3}}{2}\cdot{ \frac{\sqrt{3}}{8}}=\frac{3}{16}[/tex]