Question

Тригонометрия. Помогите, пожалуйста!

Answer 1

$1.\;\frac12\sin4\alpha(ctg\alpha-tg\alpha)=\sin2\alpha\cos2\alpha\left(\frac{\cos\alpha}{\sin\alpha}-\frac{\sin\alpha}{\cos\alpha}\right)=\\=\sin2\alpha\cos2\alpha\cdot\frac{\cos^2\alpha-\sin^2\alpha}{\sin\alpha\cos\alpha}=\sin2\alpha\cos2\alpha\cdot\frac{\cos2\alpha}{\frac12\sin2\alpha}=\\=\sin2\alpha\cos2\alpha\cdot\frac{2\cos2\alpha}{\sin2\alpha}=2\cos^22\alpha=(2\cos^22\alpha-1)+1=\cos4\alpha+1$
$2.\;a)\;\cos(\alpha-\beta)-\cos(\alpha+\beta)=(\cos\alpha\cos\beta+\sin\alpha\sin\beta)-\\-( (\cos\alpha\cos\beta-\sin\alpha\sin\beta))= \cos\alpha\cos\beta+\sin\alpha\sin\beta-\\-\cos\alpha\cos\beta+\sin\alpha\sin\beta=0$
$b)\;\frac{\sin(-\alpha)+\cos(\pi+\alpha)}{2\cos\left(\frac\pi2- \alpha \right)\cos(-\alpha)+1}=\frac{-\sin\alpha -\cos\alpha }{2\sin\alpha \cos\alpha +1}=\frac{-(\sin\alpha+\cos\alpha)}{2\sin\alpha \cos\alpha +\sin^2\alpha +\cos^2\alpha }=\\=\frac{-(\sin\alpha+\cos\alpha)}{( \sin\alpha +\cos\alpha)^2 }=-\frac{1}{ \sin\alpha +\cos\alpha }\\3.\;4\cos\frac{3\alpha}2\cos\alpha\sin\frac\alpha2=4\sin\frac\alpha2\cos\frac{3\alpha}2\cos\alpha=4\cdot\frac{\sin(-\alpha)+\sin2\alpha}2\cos\alpha=$
$=2\cdot({-\sin\alpha+\sin2\alpha})\cdot\cos\alpha=2\sin2\alpha\cos\alpha-2\sin\alpha\cos\alpha=\\=2\sin2\alpha\cos\alpha-\sin2\alpha=\sin2\alpha(2\cos\alpha-1)$
$4.\;\frac{\cos3\alpha+\cos2\alpha+\cos\alpha+1}{\cos\alpha+2\cos^2\frac\alpha2-1}=2\cos\frac{3\alpha}2\cos\frac\alpha2\\\frac{\cos3\alpha+\cos2\alpha+\cos\alpha+1}{\cos\alpha+2\cos^2\frac\alpha2-1}=\frac{4\cos^3\alpha-3\cos\alpha+2\cos^2\alpha-1+\cos\alpha+1}{\cos\alpha+\cos\alpha}=\\=\frac{4\cos^3\alpha-2\cos\alpha+2\cos^2\alpha}{2\cos\alpha}=\frac{2\cos\alpha(\cos^2\alpha-1)+2\cos^2\alpha}{2\cos\alpha}=$
$=\frac{2\cos\alpha\cos2\alpha+2\cos^2\alpha}{2\cos\alpha}=\frac{2\cos\alpha(\cos2\alpha+\cos\alpha)}{2\cos\alpha}=\cos2\alpha+\cos\alpha\\2\cos\frac{3\alpha}2\cos\frac{\alpha}2=2\cdot\frac{\cos\alpha+\cos2\alpha}2=\cos\alpha+\cos2\alpha\\{\cos\alpha+\cos2\alpha}=\cos\alpha+\cos2\alpha$