Тригонометрия Плиззззззззз 39 б

Question 1

Question 2

Решите задачу:

$4. cos47*cos17+sin47*sin17=cos(47-17)=cos30=0.5$

$5. \frac{sin(\alpha-\beta)+2cos\alpha*sin\beta}{2cos\alpha *cos\beta* cos(\alpha-\beta)}=\frac{sin\alpha*cos\beta-cos\alpha*sin\beta+2cos\alpha*sin\beta}{(cos(\alpha+\beta)+cos(\alpha-\beta))*cos(\alpha-\beta)}=\\=\frac{sin\alpha*cos\beta+cos\alpha*sin\beta}{(cos(\alpha+\beta)+cos(\alpha-\beta))*cos(\alpha-\beta)}=\frac{sin(\alpha+\beta)}{cos(\alpha+\beta)cos(\alpha-\beta)+cos^2(\alpha-\beta)}=\\=\frac{2sin(\alpha+\beta)}{cos2\alpha*cos2\beta+cos^2(\alpha-\beta)}=$
$\frac{2sin(\alpha+\beta)}{1/2(cos2(\alpha+\beta)+cos2(\alpha-\beta))+cos^2(\alpha-\beta)}=\\=\frac{2sin(\alpha+\beta)}{1/2(2cos^2(\alpha+\beta)-1+2cos^2(\alpha-\beta)-1)+cos^2(\alpha-\beta)}=\\=\frac{2sin(\alpha+\beta)}{cos^2(\alpha+\beta)+cos^2(\alpha-\beta)-1+cos^2(\alpha-\beta)}=\\=\frac{2sin(\alpha+\beta)}{2cos^2(\alpha-\beta)-sin^2(\alpha+\beta)}$

$6. \frac{sin(2\pi-\alpha)*tg(\pi/2+\alpha)-ctg(3\pi/2-\alpha)}{cos(2\pi+\alpha)*tg(\pi-\alpha)}=\frac{-sin\alpha*(-ctg\alpha)-tg\alpha}{cos\alpha*(-tg\alpha)}=\\=\frac{sin\alpha*\frac{cos\alpha}{\sin\alpha}-\frac{sin\alpha}{cos\alpha}}{-cos\alpha*\frac{sin\alpha}{cos\alpha}}=\frac{cos^2\alpha-sin\alpha}{-cos\alpha*sin\alpha}=-\frac{cos\alpha}{sin\alpha}+\frac{1}{cos\alpha}=sec\alpha-ctg\alpha$