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Mavannkas: Najprostsza postać
| π | | 3 | |
sin( |
| −α)cos(π−α)−cos(π+α)sin( |
| π+α) |
| 2 | | 2 | |
9 lut 14:37
Mavannkas: Jeszcze nie wiem jak się zabrać za to
| π | | sin(π+x) | | sin(π−x) | |
tg( |
| −x)( |
| + |
| )=2 |
| 2 | | cos(π−x) | | cos(−x) | |
9 lut 14:42
Chyba o to cho :): 1)
tg(π+α) = tg(α)
ctg(π/2 − α) = −tg(α)
A więc tg(π+α) + ctg(π/2 − α) = tg(α) − tg(α) = 0
2)
sin(π/2 − α) = cos(α)
cos(π−α) = − cos(α)
cos(π+α) = − cos(α)
sin(3/2π + α) = sin( −α) = − sin(α)
A więc: cos(α)*(−cos(α)) − (− cos(α) * sin(α)) = −cos(2)α+sin(α)cos(α) = cos(α) (sin(α) −
cos(α))
9 lut 14:56
Mila:
1)
tg(π+α)=tgα
2)
sin(π/2−α)*cos(π−α)−cos(π+α)*sin(3π/2+α)=
=cosα*(−cosα)−(−cosα*(−cosα))=
=−cos
2α−cos
2α=
=−2cos
2α
===========
9 lut 16:14