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hulovu
@hulovu
July 2022
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Cos(a+b) и Cos(a-b), если sina= 8/17, cosb=3/5, "пи"/2<a<"пи" , 3пи/2<a<2пи
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dnepr1
Verified answer
Сos(a+b) =
сos a*cos b - sin a*sin b.
Находим cos a =
√(1-(64 / 289)) = 15 / 17.
sin b = √(1-(9/25)) = 4 / 5.
cos(a+b) = (15 / 17)*(3 / 5) - (8 / 17)*(4 / 5) = (45-32) / 85 = 13 / 85
10 votes
Thanks 18
hulovu
а cos(a-b) если можно пожалуйста)
dnepr1
Там только знак меняется: сos(a-b) = сos a*cos b+sin a*sin b = (45+32) / 85 = 77/85.
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Answers & Comments
Verified answer
Сos(a+b) = сos a*cos b - sin a*sin b.Находим cos a = √(1-(64 / 289)) = 15 / 17.
sin b = √(1-(9/25)) = 4 / 5.
cos(a+b) = (15 / 17)*(3 / 5) - (8 / 17)*(4 / 5) = (45-32) / 85 = 13 / 85