Question

Решите пожалуйста, что сможете:D Срочно нужно до завтра

Answer 1

$1^0.\\ a) \sin51^0\cos6^0-\cos51^0\sin6^0=\\ |\sin\alpha\cos\beta-\cos\alpha\sin\beta=\sin(\alpha-\beta);|\\ =\sin(51^0-6^0)=\sin45^0= \frac{ \sqrt{2} }{2} ;\\ b)\cos61^0sin89^0+\sin61^0\cos89^0=\\ |\sin\alpha\cos\beta+\cos\alpha\sin\beta=\sin(\alpha+\beta)|\\ =sin(89^0+61^0)=sin(150^0)=sin(90^0+60^0)=\\ =\sin90^0\cos60^0+\cos90^0\sin60^0=1*cos60^0+0*sin60^0=\\ =cos60^0= \frac{1}{2}\\ \\ 2^0.\ \ 4\sin(4x)\tan(2x)=4*2\sin(2x)\cos(2x) \frac{\sin(2x)}{\cos(2x)}=8\sin^2(2x);$
$3^0. \ \ \sin\alpha= \frac{8}{17},\ \ \frac{\pi}{2}<\alpha<\pi; \ \alpha\in ( \frac{\pi}{2};\pi); \\ \alpha=(-1)^n\arcsin( \frac{8}{17})+\pi n, n\in Z;\\ if\ \ n=1 \ \ then\ \ \alpha=(-1)^1\arcsin( \frac{8}{17})+\pi*1=\\ =-\arcsin( \frac{8}{17})+\pi=\alpha \in (- \frac{\pi}{2}; \pi);\\ \alpha=\pi-\arcsin( \frac{8}{17});$
$4^0. \\ \frac{\sin(\phi+45^0)-\sin(\phi-45^0)}{\\sin(\phi+45^0)+\sin(\phi-45^0)}=\\ \\|\sin(\phi+45^0)-\sin(\phi-45^0)=\\ =\sin\phi\cos45^0+\sin45^0\cos\phi-\sin\phi\cos45^0+\sin45^0\cos\phi=\\ =2\sin45^0\cos\phi;\\ \sin(\phi+45^0)+\sin(\phi-45^0)=\\ =\sin\phi\cos45^0+\sin45^0\cos\phi+\sin\phi\cos45^0-\sin45^0\cos\phi=\\ =2\sin\phi\cos45^0;|\\ = \frac{2\sin45^0\cos\phi}{2\sin\phi\cos45^0}=\tan45^0 \frac{\cos\phi}{\sin\phi}=1*ctg\phi=ctg\phi;$
$5^0. \ \ ctg\alpha= \frac{\sqrt{3}}{9};\\ \sqrt{3}tg( \frac{11\pi}{6}-2\alpha)=\\ =\sqrt{3} \frac{tg( \frac{11\pi}{6} )-tg2\alpha}{1+tg(\frac{11\pi}{6})tg2\alpha}$