Question

Решите заадние по алгебре

Answer 1

4. $\cos^2(73^\circ) + \cos(47^\circ)\cos(73^\circ) + \cos^2(47^\circ) = I$

$\cos^2(\alpha) = \frac{1}{2}\cdot(1+\cos(2\alpha))$

$\cos(\alpha)\cdot\cos(\beta) = \frac{1}{2}\cdot(\cos(\alpha +\beta)+\cos(\alpha - \beta))$

$\cos^2(73^\circ) = \frac{1}{2}\cdot(1+\cos(146^\circ))$

$\cos^2(47^\circ) = \frac{1}{2}\cdot(1+\cos(94^\circ))$

$\cos(47^\circ)\cos(73^\circ) = \frac{1}{2}\cdot(\cos(120^\circ)+\cos(-26^\circ)) =$

$= \frac{1}{2}\cdot(\cos(120^\circ) + \cos(26^\circ))$

$I = \frac{1}{2}\cdot(1+\cos(146^\circ)+\cos(120^\circ) + \cos(26^\circ)+$

$+ 1+\cos(94^\circ)) = \frac{1}{2}\cdot(2+(\cos(146^\circ) + \cos(94^\circ)) +$

$+ \cos(120^\circ) + \cos(26^\circ)) = II$

$\cos x + \cos y = 2\cdot\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})$

$\cos(146^\circ) + \cos(94^\circ) = 2\cos(\frac{240^\circ}{2})\cos(\frac{52^\circ}{2}) =$

$= 2\cos(120^\circ)\cos(26^\circ)$

$II = \frac{1}{2}\cdot( 2+2\cos(120^\circ)\cos(26^\circ) + \cos(120^\circ) +$

$+ \cos(26^\circ) )= III$

$\cos(120^\circ) = \cos(180^\circ - 60^\circ) = -\cos(60^\circ) = -\frac{1}{2}$

$III = \frac{1}{2}\cdot( 2+ 2\cdot(-\frac{1}{2})\cdot\cos(26^\circ) + (-\frac{1}{2}) +$

$+ \cos(26^\circ)) = \frac{1}{2}\cdot( 2 - \cos(26^\circ) -\frac{1}{2} + \cos(26^\circ)) =$

$= \frac{1}{2}\cdot( 2 - \frac{1}{2}) = 1 - \frac{1}{4} = \frac{3}{4}$

5.

$\frac{\cos(11\alpha)+3\cos(9\alpha)+3\cos(7\alpha)+\cos(5\alpha)}{\cos(8\alpha)} = I$

$\cos(\alpha) = \frac{1}{3}$

$\cos(x) + \cos(y) = 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})$

$I = \frac{(\cos(11\alpha)+\cos(5\alpha))+3(\cos(9\alpha)+\cos(7\alpha))}{\cos(8\alpha)} =$

$= \frac{2\cos(8\alpha)\cos(3\alpha)+3\cdot 2\cos(8\alpha)\cos(\alpha)}{\cos(8\alpha)} =$

$= 2\cos(3\alpha) + 6\cos(\alpha) = 2\cos(3\alpha) + \frac{6}{3} =$

$= 2\cos(3\alpha) + 2  = II$

$\cos(3\alpha) = \cos(\alpha + 2\alpha) = \cos(\alpha)\cos(2\alpha) -$

$- \sin(\alpha)\sin(2\alpha) = \cos(\alpha)(2\cos^2(\alpha) - 1) -$

$- \sin(\alpha)\cdot 2\sin(\alpha)\cos(\alpha) =$

$= 2\cos^3(\alpha) - \cos(\alpha) - 2\cos(\alpha)\sin^2(\alpha) =$

$= 2\cos^3(\alpha) - \cos(\alpha) - 2\cos(\alpha)(1-\cos^2(\alpha)) =$

$= 2\cos^3(\alpha) - \cos(\alpha) - 2\cos(\alpha) + 2\cos^3(\alpha) =$

$= 4\cos^3(\alpha) - 3\cos(\alpha) = 4\cdot(\frac{1}{3})^3 - 3\cdot\frac{1}{3} =$

$= \frac{4}{27} - 1$

$II = 2\cdot(\frac{4}{27} - 1) + 2 = \frac{8}{27} - 2 + 2 = \frac{8}{27}$