Question

1) cos(альфа-пи)*ctg(пи/2+альфа)*sin(4пи-альфа) / sin(5пи+альфа)*ctg(3пи/2-альфа) 2) вычислить: sina(альфа), cosa, tga, если ctga=корень из 2, пи

Answer 1

$1.\frac{\cos(\alpha-\pi)\cdot ctg(\frac{\pi}2-\alpha)\cdot\sin(4\pi-\alpha)}{\sin(5\pi+\alpha)\cdot ctg(\frac{3\pi}2-\alpha)}=\frac{\cos(\pi-\alpha)\cdot ctg(\frac{\pi}2-\alpha)\cdot\sin(4\pi-\alpha)}{\sin(5\pi+\alpha)\cdot ctg(\frac{3\pi}2-\alpha)}=\\=\frac{-\cos\alpha\cdot tg\alpha\cdot(-\sin\alpha)}{-\sin\alpha\cdot tg\alpha}=-\cos\alpha$

$2.ctg\alpha=\sqrt2\\tg\alpha=\frac1{ctg\alpha}=\frac1{\sqrt2}\\1+ctg^2\alpha=\frac1{\sin^2\alpha}\Rightarrow\sin^2\alpha=\frac1{1+ctg^2\alpha}=\frac1{1+2}=\frac13\Rightarrow\sin\alpha=\frac1{\sqrt3}\\\cos\alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-\frac13}=\sqrt{\frac23}\\$

$3.\frac{ctg^2\alpha-tg^2\alpha}{1-tg^2\alpha}=\frac{\frac{\cos^2\alpha}{\sin^2\alpha}-\frac{\sin^2\alpha}{\cos^2\alpha}}{1-\frac{\sin^2\alpha}{\cos^2\alpha}}=\frac{\cos^4\alpha-\sin^4\alpha}{\sin^2\alpha\cos^2}:\frac{\cos^2\alpha-\sin^2\alpha}{\cos^2\alpha}=\\\frac{(\cos^2\alpha-\sin^2\alpha)((\cos^2\alpha+\sin^2\alpha)}{\sin^2\alpha\cos^2\alpha}\frac{\cos^2\alpha}{(\cos^2\alpha-\sin^2\alpha)}=$

$=\frac{\cos^2\alpha+\sin^2\alpha}{\sin^2\alpha}=\frac1{\sin^2\alpha}\\\sin\alpha\sin\beta=\frac{\cos{(\alpha-\beta)}-\cos{(\alpha+\beta)}}{2}\\4.\sin77\sin13=\frac{\cos64-\cos90}2=\frac{\cos64}2$