Question

6. Exprimați în funcţie de sin x sau cos x: a) sin(13pi-x), sin(100pi-x), sin(15pi+x); b) cos(17π-x), cos(20pi-x), cos(19pi+x), cos(24π + x). ​

Answer 1

Functiile sin si cos sunt functii periodice de perioada 2kπ,(k∈Z).

Adica:

$sin(x+2k\pi)=sin(x)\\cos(x+2k\pi)=cos(x)$

Functia sin este impara:

$sin(-x)=-sin(x)$

Functia cos este para:

$cos(-x)=cos(x)$

Alte informatii utile:

$sin(\pi+x)=sin(\pi)cos(x)+sin(x)cos(\pi)=0*cos(x)+sin(x)*(-1)=-sin(x)$

$sin(\pi-x)=sin(\pi)cos(x)-sin(x)cos(\pi)=0*cos(x)-sin(x)*(-1)=sin(x)$

$cos(\pi+x)=cos(\pi)cos(x)-sin(\pi)sin(x)=(-1)*cos(x)-0*sin(x)=-cos(x)$

$cos(\pi-x)=cos(\pi)cos(x)+sin(\pi)sin(x)=(-1)*cos(x)+0*sin(x)=-cos(x)$

a)

$sin(13\pi-x)=sin(2*6\pi+\pi-x)=sin(\pi-x)=sin(x)$

$sin(100\pi-x)=sin(2*50\pi-x)=sin(-x)=-sin(x)$

$sin(15\pi+x)=sin(2*7\pi+\pi+x)=sin(\pi+x)=-sin(x)$

b)

$cos(17\pi-x)=cos(2*8\pi+\pi-x)=cos(\pi-x)=-cos(x)$

$cos(20\pi-x)=cos(2*10\pi-x)=cos(-x)=cos(x)$

$cos(19\pi+x)=cos(2*9\pi+\pi+x)=cos(\pi+x)=-cos(x)$

$cos(24\pi+x)=cos(2*12\pi+x)=cos(x)$