Question

Срочно нужна помощь !!!

Answer 1

$\frac{cos( \frac{11\pi}{20} )cos( \frac{9\pi}{20} ) -sin( \frac{11\pi}{20} )sin( \frac{9\pi}{20} ) }{sin( \frac{\pi}{8} ) cos( \frac{3\pi}{8} ) -cos( \frac{\pi}{8} )sin( \frac{3\pi}{8} ) } = \frac{cos( \frac{11\pi}{20} + \frac{9\pi}{20} )}{sin( \frac{\pi}{8}- \frac{3\pi}{8} ) } = \frac{cos(\pi)}{sin(- \frac{\pi}{4} )} = \frac{2}{ \sqrt{2} } = \sqrt{2\\} \\.................................$
$\frac{1+tg(1)tg(46)}{tg(1)-tg(46)} = \frac{ \frac{1}{tg(1)-tg(46)} }{1+tg(1)tg(46)} = \frac{1}{tg(-45)} =-1$
Там градусы у тангинсах 
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$sin(2a)+cos(2a)ctg(a)=2sin(a)cos(a)+(cos^2(a)-sin^2(a))ctg(a)=\\=2sin(a)cos(a)+ \frac{cos(a)cos^2(a)}{sin(a)} - \frac{sin^2(a)cos(a)}{sin(a)} =2sin(a)cos(a)+\\+ \frac{cos(a)cos(2a)}{sin(a)} =sin(2x)sin(x)+cos(a)cos(2a)=cos(-a)=cos(a)$
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$ctg(2a)-tg(2a)= \frac{1}{tg(2a)} -tg(2a)= \frac{1-tg^2(2a)}{tg(2a)} = \frac{1}{ \frac{1}{2}tg(4a) } = \frac{ \frac{1}{tg(4a)} }{2} = \\= \frac{2}{tg(4a)}$