Ответы на вопрос:
Ctg a = 3; ctg²a = 9; sin²a = 1/(1 + ctg²a) = 1/(1 + 9) = 1/10 cos(2a-5пи) = cos(-5пи + 2a) = cos(5пи - 2a) = -cos2a = - (1 - 2sin²a) =2sin²a - 1 = 2*1/10 -1 = 2/10 - 1 = 1/5 - 1 = - 4/5
Cos(2a-5pi)=-cos2a=1-2cos²a=1-2/(1+tg²a) tga=1/ctga=1/3 cos(2a-5pi)=1-2*9/10=1-9/5=-4/5
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