1) известно что:2) упростите выражения 3) докажите тождество

Question 1

1) известно что:
2) упростите выражения
3) докажите тождество

Question 2

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Question 3

Решите задачу:

$sin(\frac{\pi}{2}+a)=-\frac{1}{2}\\ sin(30+a)\\ \\ sin(\frac{\pi}{2}+a)=cosa\\ cosa=-\frac{1}{2}\\ a=-\frac{2\pi}{3}+2\pi*n=\frac{4pi}{3}\\\\ sin(30+240)=sin(270)=-1\\\\ 2cos^2a-(tga*cosa)^2-(ctga*sina)^2=\\ 2cos^2a-(\frac{sina}{cosa}*cosa)^2-(\frac{cosa}{sina}*sina)^2=\\ 2cos^2a-sin^2a-cos^2a=\\ cos^2a-sin^2a=cos2a\\ \\$

$\frac{sina+sin3a}{cosa+cos3a}*(1+cos4a)=\\ \frac{2sin2a*cosa}{2cos2a*cosa}*(1+cos4a)=\\ \frac{sin2a}{cos2a}*(1+cos(2*2a))=\\ \frac{sin2a}{cos2a}*(2cos^2(2a))=\\ 2sin2a*cos2a=sin4a\\ \\ ctg2a*\frac{2tga}{1+tg^2a}=\\ ctg2a*sin2a=\\ \frac{cos2a}{sin2a}*sin2a=cos2a$

Question 4

1)
$\sin(\frac{\pi}{2}+\alpha)=-\frac{1}{2}; \\ \pi<\alpha<\frac{3\pi}{2};\ \alpha\in(\pi;\frac{2\pi}{3});\\ \sin(30^0+\alpha)-?;\\ \frac{\pi}{2}+\alpha=(-1)^n\arcsin(-\frac{1}{2})+\pi n;\ n\in Z\\ \alpha=-(-1)^n\arcsin(\frac{1}{2})-\frac{\pi}{2}+\pi n;\\ \alpha=(-1)^{n+1}\cdot\frac{\pi}{6}-\frac{\pi}{2}+\pi n;\\ n=2: \ \alpha=-\frac{\pi}{6}-\frac{\pi}{2}+2\pi=\frac{-\pi-3\pi+12\pi}{6}=\frac{8\pi}{6}=\frac{4\pi}{3}\in(\pi;\frac{2\pi}{3});\\$
$\sin(30^0+\alpha)=\sin(\frac{\pi}{6}+\frac{4\pi}{3})=\sin(\frac{\pi+8\pi}{6})=\\ =\sin\frac{9\pi}{6}=\sin{\frac{3\pi}{2}}-1;\\ or: \sin\frac{\pi}{6}\cos\frac{4\pi}{3}+\sin{\frac{4\pi}{3}}\cos\frac{\pi}{6}=\\ =\frac{1}{2}\cdot(-\frac{1}{2})+(-\frac{\sqrt{3}}{2})\cdot\frac{\sqrt{3}}{2}=-\frac{1}{4}-\frac{3}{4}=-\frac{4}{4}=-1$

$\sin(30^0+\alpha)=-1;\\$

2)
a)
$2\cos^2\alpha-(tg\alpha\cdot\cos\alpha)^2-(ctg\alpha\cdot\sin\alpha)^2=\\ =2\cos^2\alpha-\left(\left(\frac{\sin\alpha}{\cos\alpha}\cdot\cos\alpha\right)^2+\left(\frac{\cos\alpha}{\sin\alpha}\cdot sin\alpha\right)^2\right)=\\ =2\cos^2\alpha-\left(\sin^2\alpha+\cos^2\alpha\right)=2\cos^2\alpha-1=\cos2\alpha;\\$
б)
$\frac{\sin\alpha+\sin3\alpha}{\cos\alpha+\cos3\alpha}\cdot(1+\cos4\alpha)=\\ =\frac{\sin\alpha+\sin(\alpha+2\alpha)}{\cos\alpha+\cos(\alpha+2\alpha)}\cdot(1+2\cos^22\alpha-1)=\\ =\frac{\sin\alpha+\sin\alpha\cos2\alpha+\cos\alpha\sin2\alpha}{\cos\alpha+\cos\alpha\cos2\alpha-\sin\alpha\sin2\alpha}\cdot2\cos^22\alpha=\\ ==\frac{\sin\alpha+\sin\alpha\cos2\alpha+2\cos^2\alpha\sin\alpha}{\cos\alpha+\cos\alpha\cos2\alpha-2\sin^2\alpha\cos\alpha}\cdot2\cos^22\alpha=\\$
$=\frac{\sin\alpha(1+\cos2\alpha+2\cos^2\alpha)}{\cos\alpha(1+\cos2\alpha-2\sin^2\alpha)}\cdot2\cos^22\alpha=\\ \frac{\sin\alpha(1+2\cos^2\alpha-1+2\cos^2\alpha)}{\cos\alpha(\cos2\alpha+1-2\sin^2\alpha)}\cdot2\cos^22\alpha=\\ =\frac{\sin\alpha\cdot4\cos^2\alpha}{\cos\alpha\cdot2\cos2\alpha}\cdot2\cos^22\alpha=\\ =4\sin\alpha\cos\alpha\cos2\alpha=2\sin2\alpha2\cos2\alpha=\sin4\alpha;\\$
3)
$ctg2\alpha\frac{2tg\alpha}{1+tg^2\alpha}=\cos2\alpha;\\ \frac{\cos2\alpha}{\sin2\alpha}\cdot\frac{2\frac{sin\alpha}{\cos\alpha}}{1+(\frac{\sin\alpha}{\cos\alpha})^2}=\frac{\cos2\alpha}{\sin2\alpha}\cdot\frac{2\frac{\sin\alpha}{\cos\alpha}}{\frac{\cos^2\alpha+\sin^2\alpha}{\cos^2\alpha}}=2\cdot\frac{\cos2\alpha}{\sin2\alpha}\cdot\frac{\frac{\sin\alpha}{\cos\alpha}}{\frac{1}{\cos^2\alpha}}=\\ =2\cdot\frac{\cos2\alpha}{\sin2\alpha}\cdot\frac{\sin\alpha\cdot\cos^2\alpha}{\cos\alpha}=$
$=2\cdot\frac{\cos2\alpha}{\sin2\alpha}\cdot\sin\alpha\cdot\cos\alpha=\frac{\cos2\alpha}{\sin2\alpha}\cdot2\sin\alpha\cos\alpha=\frac{\cos2\alpha}{\sin2\alpha}\cdot\sin2\alpha=\cos2\alpha;$