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Question 1

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Question 2

Решите задачу:

$1^o\\a)\;\sin113^o\cos67^o+\sin67^o\cos113^o=\sin(113^o+67^o)=\sin180^o=0\\b)\;\cos74^o\cos29^o+\sin74^o\cos61^o=\frac12\left(\cos45^o+\cos103^o+\sin13^o+\sin135^o\right)=\;=\frac12\left(\cos45^o+\sin135^o+\cos(\frac\pi2+13^o)+\sin13^o\right)=\\=\frac12\left(\frac{\sqrt2}2+\frac{\sqrt2}2-\sin13^o+\sin13^o\right)=\frac{\sqrt2}2$

$2^o\\2\cos^23\alpha tg3\alpha=5\cos^23\alpha\cdot\frac{\sin3\alpha}{\cos\3\alpha}=5\cos3\alpha\sin3\alpha=2,5\cdot\sin6\alpha\\\\3.\\\sin^2\alpha+\cos^\alpha=1\Rightarrow\cos\alpha=\sqrt{1-\sin^2\alpha}\\\cos\alpha=\sqrt{1-0,64}=\sqrt{0,36}=\pm0,6\\\frac\pi2<\alpha\pi\Rightarrow\cos\alpha<0\\\cos\alpha=-0,6$

$4.\\\frac{\sin(x+45^o)+\sin(x-45^o)}{\sin(x+45^o)-\sin(x-45^o)}=\frac{\sin x\cos45^o+\cos x\sin45^o+\sin x\cos45^o-\cos x\sin45^o}{\sin x\cos45^o+\cos x\sin45^o-\sin x\cos45^o+\cos x\sin45^o}=\\=\frac{2\sin x\cos45^o}{2\cos x\sin45^o}=\frac{\sqrt2\sin x}{\sqrt2\cos x}=tgx$

$5.\\\cos\alpha=\frac23\\\sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-\frac49}=\sqrt{\frac59}=\pm\frac{\sqrt5}3\\270^o<\alpha<360^o\Rightarrow\sin\alpha<0\\\sin\alpha=-\frac{\sqrt5}3\\\sin\beta=\frac13\\\cos\beta=\sqrt{1-\frac19}=\sqrt{\frac89}=\pm\frac{2\sqrt2}3\\90^o<\beta<180^o\Rightarrow\cos\beta<0\\\cos\beta=$

$\sin(2\alpha+\beta)=\sin2\alpha\cos\beta+\cos2\alpha\sin\beta=\\=2\sin\alpha\cos\alpha\cos\beta+(1-\sin^2\alpha)\sin\beta=\\=2\cdot\left(-\frac{\sqrt5}3\right)\cdot\frac23\cdot\left(-\frac{2\sqrt2}3\right)+\left(1-\frac59\right)\cdot\frac13=\frac{8\sqrt{10}+4}{27}$