Question

50 .только с нормальным решением

Answer 1

$1)\;\sin\alpha\sin(\alpha+\beta)+\cos\alpha\cos(\alpha+\beta)=\\=\sin\alpha(\sin\alpha\cos\beta+\cos\alpha\sin\beta)+\cos\alpha(\cos\alpha\cos\beta-\sin\alpha\sin\beta)=\\=\sin^2\alpha\cos\beta+\sin\alpha\cos\alpha\sin\beta+\cos^2\alpha\sin\beta-\sin\alpha\cos\alpha\sin\beta=\\=\sin^2\alpha\cos\beta+\cos^2\alpha\sin\beta$

$2)\;\frac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}{\cos(\alpha-\beta)-\cos(\alpha+\beta)}=\frac{\cos\alpha\cos\beta-\sin\alpha\sin\beta+\cos\alpha\cos\beta+\sin\alpha\sin\beta}{\cos\alpha\cos\beta+\sin\alpha\sin\beta-\cos\alpha\cos\beta+\sin\alpha\sin\beta}=\\\\=\frac{2\cos\alpha\cos\beta}{2\sin\alpha\sin\beta}=ctg\alpha\cdot ctg\beta$

$3)\;\frac{\sin11^o\cos15^o+\cos11^o\sin15^o}{\sin18^o\cos12^o+\cos18^o\sin12^o}=\frac{\sin(11^o+15^o)}{\sin(18^o+12^o)}=\frac{\sin26^o}{\sin30^o}=2\sin26^o$

$4)\;\frac{tg7\alpha-tg3\alpha}{tg7\alpha\cdot tg3\alpha+1}=tg(7\alpha-3\alpha)=tg4\alpha\\\\5)\;\cos^2\alpha-4\sin^2\frac\alpha2\cos^2\frac\alpha2=\cos^2\alpha-(2\sin\frac\alpha2\cos\frac\alpha2)^2=\cos^2\alpha-(\sin\alpha)^2=\\=\cos^2\alpha-\sin^2\alpha=\cos2\alpha$

$6)\;\sqrt{1+\cos8\alpha}=\sqrt{1+2\cos^24\alpha-1}=\sqrt{2\cos^24\alpha}=\sqrt2\cdot\cos4\alpha\\\\7)\;\frac{\sin\alpha\cos\alpha}{1-2\sin^2\alpha}=\frac{\frac12\sin2\alpha}{\cos2\alpha}=\frac12tg2\alpha$