Решение
1. а) cos²π/8 - sin²π/8 = cos(2*π/8) = cosπ/4 = √2/2
б) sinπ/7cosπ/21 + cosπ/7sinπ/21 = sin(π/7 + π/21) = sin(7π/21) sin(π/3) = √3/2
3. f(x) = cos2x - 3sinx
а) f(0) = cos0 - 3sin0 = 1 - 3*0 = 1
б) f(π/2) = cosπ - 3sin(π/2) = - 1 - 3*1 = - 4
в) f(π/6) = cos(π/3) - 3sin(π/6) = 1/2 - 3*(1/2) = - 3
5. 2cos2x + cosx + 0,5 = 0
2*(2cos²x - 1) + cosx + 0,5 = 0
4cos²x - 2 + cosx + 0,5 = 0
cosx = t, I tI ≥ 1
4t² + t - 1,5 = 0
D = 1 + 4*4*1,5 = 25
t1 = (-1 - 5)/8
t1 = - 3/4
t2 = (- 1 + 5)/8
t2 = 1/2
1) cosx = - 3/4
x = (+ -)arccos(-3/4) + 2πk, k∈Z
2) cosx = 1/2
x = (+ -)arccos(1/2) + 2πn, n∈Z
x = (+ -)π/3 + 2πn, n∈Z