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  • Użytkownik Zadane
2009-11-19T22:54:09+01:00
[cos³α-cosα]÷[sin³α-sinα]=tagα
L = cos³α-cos α÷sin³α-sinα
P = tagα
L = [cos³α-cosα]÷[sin³α-sinα] = [cosα (cos²α - 1)] : [sinα (sin²α -1)] =
[cosα (1 -sin²α - 1)] : [sinα (1 - cos²α - 1)] =
[cosα*(-sin²α)] : [sinα*(-cos²α)] =
[cosα*sin²α] : [sinα*cos²α] =
[sinα] : [cosα] =
sinα/cosα = tagα = P