Question

Кто-нибудь справиться с домахой нужны не только ответы, но и решение буду премного

Answer 1

$1)\; \; cos54^\circ \, cos6^\circ -sin54^\circ \, sin6^\circ =cos(54^\circ +6^\circ )=cos60^\circ =\frac{1}{2}\\\\cos\frac{3\pi}{10}\, cos\frac{\pi}{20}+sin\frac{\pi}{20}\, sin\frac{3\pi}{10}=cos(\frac{3\pi}{10}-\frac{\pi}{20})=cos\frac{\pi}{4}=\frac{\sqrt2}{2}$

$2)\; \; sin(3x+2y)\, cos(x+2y)-sin(x+2y)\, cos(3x+2y)=\\\\=sin(3x+2y-x-2y)=sin2x\\\\3)\; \; \frac{sin13^\circ \, cos47^\circ +sin47^\circ \, cos13^\circ }{cis98^\circ \, cos38^\circ +sin98^\circ \, sin38^\circ }=\frac{sin(13^\circ +47^\circ )}{cos(98^\circ -38^\circ )}=\frac{sin60^\circ }{cos60^\circ }=tg60^\circ =\sqrt3$

$4)\; \; \frac{sin58^\circ \, cos52^\circ +sin52^\circ \, cos58^\circ }{cos68^\circ \, cos42^\circ -sin42^\circ \, sin68^\circ }=\frac{sin(58^\circ +52^\circ )}{cos(68^\circ +42^\circ )}=\frac{sin110^\circ }{cos110^\circ }=\\\\=\frac{sin(90^\circ +20^\circ )}{cos(90^\circ +20^\circ )}=\frac{cos20^\circ }{-sin20^\circ }=-ctg20^\circ <0\\\\\frac{sin48^\circ +cos48^\circ }{cos24^\circ -sin24^\circ }=\frac{sin48^\circ +cos(90^\circ -42^\circ )}{cos24^\circ -sin(90^\circ -66^\circ )}=\frac{sin48^\circ +sin42^\circ }{cos24^\circ -cos66^\circ }=$

$=\frac{2\, sin45^\circ \, cos3^\circ }{2\, sin45^\circ \, sin42^\circ }=\frac{cos3^\circ }{sin42^\circ }0\\\\\frac{cos3^\circ }{sin42^\circ }-ctg20^\circ \\\\5)\; \; sinx+cosx=\sqrt2\cdot (\frac{1}{\sqrt2}\cdot sinx+\frac{1}{\sqrt2}\cdot cosx)=\\\\=\sqrt2\cdot (cos\frac{\pi}{4}\cdot sinx+sin\frac{\pi}{4}\cdot cosx)=\sqrt2\cdot sin(x+\frac{\pi}{4})\\\\-1\leq sin(x+\frac{\pi}{4})\leq 1\\\\-\sqrt2\; \leq \sqrt2\cdot sin(x+\frac{\pi}{4})\leq \; \sqrt2$

Наименьшее значение  $-\sqrt2$  , наибольшее  $\sqrt2$  .

$6)\; \; sin(arcsin0,6+arcsun\frac{12}{13})=\\\\=sin(arcsin0,6)\cdot cos(arcsin\frac{12}{13})+cos(arcsin0,6)\cdot sin(arcsin\frac{12}{13})=\\\\=0,6\cdot \frac{5}{13}+0,8\cdot \frac{12}{13}=\frac{3\cdot 5+4\cdot 12}{5\cdot 13}=\frac{63}{65}$

Answer 2

1) а) cos(54+6)=cos(60)

б)

$\cos( \frac{3\pi}{10} - \frac{\pi}{20} ) = \cos( \frac{\pi}{4} ) = \frac{ \sqrt{2} }{2}$

2)

$\sin(3x + 2y - x - 2y) = \sin(2x)$

3)

$\frac{ \sin(60) }{ \cos(60) } = \tan(60)$

4)

$\frac{ \sin(110) }{ \cos(110) } = \tan(110)$

0,044286 > - 1,05914

5) -

6) -