Ответы
Ответ дал:
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a) [2cos(2π/3 + π/6)/2] * [cos(2π/3 - π/6)/2] = 2cos(5π/12) * cos(π/4)
б) [2cos(3π/10 + π/10)/2] * [sin(3π/10 - π/10)/2] =
= 2cos(π/5) * sin(π/10)
в) cosβ - cos6β = [2sin(β + 6β)/2 * sin(6β - β)/2] =
= 2sin (7β/2) * sin(5β/2)]
б) [2cos(3π/10 + π/10)/2] * [sin(3π/10 - π/10)/2] =
= 2cos(π/5) * sin(π/10)
в) cosβ - cos6β = [2sin(β + 6β)/2 * sin(6β - β)/2] =
= 2sin (7β/2) * sin(5β/2)]
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