Question

1. sin20*cos25 + cos65*cos20
2. cos8II/15 * sin II/5 - sin8II/15*cosII/5
3. sin II/12 * sin5II/12 =1/4
4. sinx-cosx = radical 2
sin2x=? ​

Answer 1

Răspuns:

Explicație pas cu pas:

1) a =sin20*cos25 + cos65*cos20 =

=1/2(sin45 +sin(-5)) +1/2(cos85 +cos45))

sin(-5) = -sin5,  cos85 = sin5 (unghiuri compl.)

a = 1/2*√2/2 +1/2*√2/2 = 2*√2/4 = √2/2

2) a = -(sin96*cos36 -sin36*cos96) =

 = -sin(96 -36) = -sin60 = -√3/2

3) sin75*sin15 = 1/2(cos60 - cos90) =

  = 1/2*1/2 = 1/4  (cos90 = 0)

4) sinx -sin(x -pi/4) = 2sin((x -pi/4 +x)/2)*cos((x +pi2/2 -x)/2) =

= 2sin(x-pi/4) *√2/2 = -√2sin(pi/4 -x)

-√2sin(pi/4 -x) =√2

-sin(pi/4 -x) = -1 = sin(-pi/2)

sin(pi/4 -x)  = sin(-pi/2)

pi/4 -x = -pi/2

x = pi/2 +pi4 +2k*pi = 3*pi/4 +2k*pi