Question

1-6

Осталось 35 минут

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Answer 1

$1)\;\frac{\cos^26\alpha-1}{1-\sin^26\alpha}-tg12\alpha ctg12\alpha=\frac{\cos^26\alpha-\sin^26\alpha-\cos^26\alpha}{\sin^26\alpha+\cos^26\alpha-\sin^26\alpha}-1=\frac{-\sin^26\alpha}{\cos^26\alpha}-1=\\=\frac{-\sin^26\alpha-\cos^26\alpha}{\cos^26\alpha}=-\frac1{\cos^26\alpha}\\\\2)\;\sin8\alpha\cos3\alpha-\cos8\alpha\sin3\alpha=\sin(8\alpha-3\alpha)=\sin5\alpha\\\\3)\;\frac{4\cos^27\alpha}{\sin14\alpha}=\frac{4\cos^27\alpha}{2\sin7\alpha\cos7\alpha}=\frac{2\cos7\alpha}{\sin7\alpha}=2ctg7\alpha$

$4)\;\frac{\sin14\alpha-\sin10\alpha}{\cos3\alpha-\cos7\alpha}=\frac{2\sin\frac{14\alpha+10\alpha}2\cos\frac{14\alpha-10\alpha}2}{-2\sin\frac{3\alpha+7\alpha}2\sin\frac{3\alpha-7\alpha}2}=\frac{\sin2\alpha\cos12\alpha}{\sin5\alpha\sin2\alpha}=\frac{\cos12\alpha}{\sin5\alpha}\\\\5)\;\cos^2\left(\frac\pi2-3\alpha\right)-\cos^2(\pi+3\alpha)=\sin^23\alpha-\cos^23\alpha=-(\cos^23\alpha-\sin^23\alpha)=\\=-\cos6\alpha$