Question

URGENTT!!!
ex 7 b va roggg

Answer 1

Explicație pas cu pas:

a)

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

$= > \sin(x)\cos(y) = \frac{1}{2} \left[ \sin(x + y) +\sin(x - y) \right]$

b)

$\sin( \frac{x}{2}) \left[ \cos(x) + \cos(2x) + \cos(3x) \right] =$

$= \frac{1}{2} \left[ \sin( \frac{x}{2} + x) + \sin( \frac{x}{2} - x) + \sin( \frac{x}{2} + 2x) + \sin( \frac{x}{2} - 2x) + \sin( \frac{x}{2} + 3x) + \sin( \frac{x}{2} - 3x)\right]$

$= \frac{1}{2} \left[ \sin( \frac{3x}{2}) + \sin( - \frac{x}{2}) + \sin( \frac{5x}{2}) + \sin( - \frac{3x}{2}) + \sin( \frac{7x}{2}) + \sin( - \frac{5x}{2})\right]$

$= \frac{1}{2} \left[ \sin( \frac{3x}{2}) - \sin( \frac{x}{2}) + \sin( \frac{5x}{2}) - \sin(\frac{3x}{2}) + \sin( \frac{7x}{2}) - \sin(\frac{5x}{2})\right]$

$= \frac{1}{2} \left[ \sin( \frac{7x}{2}) - \sin( \frac{x}{2})\right]$

c)

$S = \cos( \frac{2\pi}{9}) + \cos(\frac{4\pi}{9}) + \cos(\frac{6\pi}{9}) + \cos(\frac{8\pi}{9})$

$\sin( \frac{\pi}{9} ) \times S = \sin( \frac{\pi}{9} )\left[ \cos( \frac{2\pi}{9}) + \cos(\frac{4\pi}{9}) + \cos(\frac{6\pi}{9}) + \cos(\frac{8\pi}{9})\right]$

$= \frac{1}{2} \left[ \sin(\frac{\pi}{9} + \frac{2\pi}{9}) + \sin(\frac{\pi}{9} - \frac{2\pi}{9}) +\sin(\frac{\pi}{9} + \frac{4\pi}{9}) + \sin(\frac{\pi}{9} - \frac{4\pi}{9}) +\sin(\frac{\pi}{9} + \frac{6\pi}{9}) + \sin(\frac{\pi}{9} - \frac{6\pi}{9}) +\sin(\frac{\pi}{9} + \frac{8\pi}{9}) + \sin(\frac{\pi}{9} - \frac{8\pi}{9}) \right]$

$= \frac{1}{2} \left[ \sin(\frac{3\pi}{9}) + \sin( - \frac{\pi}{9}) +\sin(\frac{5\pi}{9}) + \sin( - \frac{3\pi}{9}) +\sin(\frac{7\pi}{9}) + \sin( - \frac{5\pi}{9}) +\sin(\frac{9\pi}{9}) + \sin( - \frac{7\pi}{9}) \right]$

$= \frac{1}{2} \left[ \sin(\frac{3\pi}{9}) - \sin(\frac{\pi}{9}) +\sin(\frac{5\pi}{9}) - \sin( \frac{3\pi}{9}) +\sin(\frac{7\pi}{9}) - \sin(\frac{5\pi}{9}) +\sin(\frac{9\pi}{9}) - \sin(\frac{7\pi}{9}) \right]$

$= \frac{1}{2} \left[\sin(\pi) - \sin(\frac{\pi}{9})\right]$

$= > S = \frac{ \left[\sin(\pi) - \sin(\frac{\pi}{9})\right]}{2\sin( \frac{\pi}{9} )}$

d)

$S = \cos(x) + \cos(2x) + \cos(3x) + ... + \cos(nx)$

$S \times 2\sin( \frac{x}{2}) = 2\sin( \frac{x}{2})\left[ \cos(x) + \cos(2x) + \cos(3x) + ... + \cos(nx) \right]$

$= \sin( \frac{x}{2} + x) + \sin( \frac{x}{2} - x) + \sin( \frac{x}{2} + 2x) + \sin( \frac{x}{2} - 2x) + \sin( \frac{x}{2} + 3x) + \sin( \frac{x}{2} - 3x) + ... + \sin( \frac{x}{2} + nx) + \sin( \frac{x}{2} - nx)$

$=\sin( \frac{3x}{2}) + \sin( - \frac{x}{2}) + \sin( \frac{5x}{2}) + \sin(- \frac{3x}{2}) + \sin( \frac{7x}{2}) + \sin(-\frac{5x}{2}) + ... + \sin( \frac{(2n + 1)x}{2}) + \sin( - \frac{(2n - 1)x}{2})$

$= \sin( \frac{3x}{2}) - \sin(\frac{x}{2}) + \sin( \frac{5x}{2}) - \sin(\frac{3x}{2}) + \sin( \frac{7x}{2}) - \sin(\frac{5x}{2}) + ... + \sin( \frac{(2n + 1)x}{2}) - \sin( \frac{(2n - 1)x}{2})$

$=\sin( \frac{(2n + 1)x}{2}) - \sin(\frac{x}{2})$

$S \times 2\sin( \frac{x}{2}) = \sin( \frac{(2n + 1)x}{2}) - \sin(\frac{x}{2})$

$= > S = \frac{\sin( \frac{(2n + 1)x}{2}) - \sin(\frac{x}{2})}{2\sin( \frac{x}{2})}$

e)

$S = \cos(a) + \cos(a + r) + \cos(a + 2r) + ... + \cos\left[ a + (n - 1)r \right]$

$S \times 2\sin( \frac{r}{2}) =2\sin( \frac{r}{2})\left[ \cos(a) + \cos(a + r) + \cos(a + 2r) + ... + \cos\left[ a + (n - 1)r \right] \right]$

$= 2\sin( \frac{r}{2})\left[ \cos(r + a) + \cos(r - a) + \cos(r + a + r) + \cos(r - a - r) + \cos(r + a + 2r) + \cos(r - a - 2r) + ... + \cos\left[ r + a + (n - 1)r \right] + \cos\left[ r - a - (n - 1)r \right]\right]$

$= \sin(\frac{r}{2} + a) + \sin(\frac{r}{2} - a) + \sin(\frac{r}{2} + a + r) + \sin(\frac{r}{2} - a - r) + \sin(\frac{r}{2} + a + 2r) + \sin(\frac{r}{2} - a - 2r) + ... + \sin\left[\frac{r}{2} + a + (n - 1)r \right] + \sin\left[\frac{r}{2} - a - (n - 1)r \right]$

$= \sin(a + \frac{r}{2}) - \sin(a - \frac{r}{2}) + \sin(a + \frac{3r}{2}) - \sin(a + \frac{r}{2}) + \sin(a + \frac{5r}{2}) - \sin(a + \frac{3r}{2}) + ... + \sin\left[a + \frac{(2n - 1)r}{2} \right] - \sin\left[a + \frac{(2n - 3)r}{2} \right]$

$= \sin\left[a + \frac{(2n - 1)r}{2} \right] - \sin(a - \frac{r}{2})$

$S \times 2\sin( \frac{r}{2}) = \sin\left[a + \frac{(2n - 1)r}{2} \right] - \sin(a - \frac{r}{2})$

$= > S = \frac{\sin\left[a + \frac{(2n - 1)r}{2} \right] - \sin(a - \frac{r}{2})}{2\sin( \frac{r}{2})}$