Question

-2cos(360-x)sinx при х=15

Answer 1

1) Вычислить:

2*cos(5x)*sin(7x)-sin(12x)=sin(2x)+sin(12x)-sim(12x)=sin(2x)=2*sin(x)*cos(x)

sinx-cosx=0,4 - >

(sin(x))^2-2*sin(x)*cos(x)+(cos(x))^2=0,16

2*sin(x)*cos(x)=(sin(x))^2+(cos(x))^2-0,16=0,84

 

2) а)

2*sin(x)=2*cos(x)+sqrt(6)

2*sin(x)-2*cos(x)=sqrt(6)

sin(x)/sqrt(2)-cos(x)/sqrt(2)=sqrt(3)/2

(1/sqrt(2))^2+1/sqrt(2))^2=1 - > 1/sqrt(2)=sin(a)=cos(a) и a=arcsin(sqrt(2)/2)=п/4

sin(x)*cos(a)-cos(x)*sin(a)=sqrt(3)/2

sin(x-a)=sqrt(3)/2

x-a=(-1)^k*п/3+пк

x=(-1)^k*п/3+п/4+пк kЄZ

Answer 2

-2cos(360-x)sinx=-2cosx*sinx=-sin2x=-sin2*15=-sin30=-0.5