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Question 1

Question 2

Решите задачу:

$1.\\a)\;\frac{tg\frac\pi3}{ctg\frac\pi3}-\sqrt2\cos\frac{3\pi}4=\frac{\frac1{\sqrt3}}{\frac1{\sqrt3}}-\sqrt2\cdot\left(-\frac1{\sqrt2}\right)=1+1=2\\b)\;\frac{\sin50^o+\sin10^o}{\cos25^o\cos5^o+\sin25^o\sin5^o}=\frac{2\sin\frac{50^o+10^o}2\cos\frac{50^o-10^o}2}{\cos(25^o-5^0)}=\frac{2\sin30^o\cos20^o}{\cos20^o}=2\cdot\frac12=1$

$2.\;\cos\left(\frac{3\pi}2+\alpha\right)=\sin\alpha=0,5\\\frac\pi2<\alpha<\pi\Rightarrow\alpha=\frac{5\pi}6\\\sin\left(60^o+\frac{5\pi}6\right)=\sin\left(\frac\pi3+\frac{5\pi}6\right)=\sin\frac{7\pi}6=-\frac12$

$3.\\a)\;\left(\frac{\sin\alpha}{tg\alpha}\right)^2+\left(\frac{\cos\alpha}{ctg\alpha}\right)^2-2\sin^2\alpha=\left(\frac{\sin\alpha}{\frac{\sin\alpha}{\cos\alpha}}\right)^2+\left(\frac{\cos\alpha}{\frac{\sin\alpha}{\cos\alpha}}\right)^2-2\sin^2\alpha=\\=\cos^2\alpha+\sin^2\alpha-2\sin^2\alpha=\cos^2\alpha-\sin^2\alpha=\cos2\alpha\$

$b)\;\frac{\sin\alpha-\sin3\alpha}{\cos\alpha-\cos3\alpha}\cdot(1-\cos4\alpha)=\frac{2\sin\frac{\alpha-3\alpha}2\cos\frac{\alpha+3\alpha}2}{-2\sin\frac{\alpha+3\alpha}2\sin\frac{\alpha-3\alpha}2}\cdot(1-(1-2\sin^22\alpha))=\\=\frac{2\sin(-\alpha)\cos2\alpha}{-2\sin2\alpha\sin(-\alpha)}\cdot(1-1+2\sin^22\alpha)=-\frac{\cos2\alpha}{\sin2\alpha}\cdot2\sin^22\alpha=-2\sin2\alpha\cos2\alpha=\\=\sin4\alpha$

$5.\;tg2\alpha\cdot\frac{1-tg^2\alpha}{1+tg^2\alpha}=tg2\alpha\cdot\cos2\alpha=\frac{\sin2\alpha}{\cos2\alpha}\cdot\cos2\alpha=\sin2\alpha$