p{(1+1/tg^{2}α)}
Ardox: Ile to jest:
√(1+1/tg2α)
16 lis 23:30
pigor: ...,
| | 1 | |
√1+1/tg2α=√1+ctg2α= ...= |
| i sinx≠0 . ...  |
| | |sinx| | |
17 lis 00:06
Dziadek Mróz:
√1 + 1/tg2(x) =
√1 + ctg2(x) =
√1 + cos2(x)/sin2(x) =
=
√sin2(x)/sin2(x) + cos2(x)/sin2(x) =
√(sin2(x) + cos2(x))/sin2(x) =
| | 1 | | 1 | |
= √1/sin2(x) = |
| = |
| |
| | √sin2(x) | | |sin(x)| | |
17 lis 00:46