wyznacz a
shersken: wyznacz liczbę a=(sin30 x sin50)/5sin100 bez pomocy tablic. Wie ktos jak sie za to zabrac?
5 mar 21:39
Basia:
cosx − cosy = −2sin
x+y2*sin
x−y2
x+y2 = 50
x−y2 = 30
x+y = 100
x−y = 60
−−−−−−−−−−−−−−
2x = 180
x = 90
y = 10
−2sin30*sin50 = cos90 − cos10
| | cos90−cos10 | | 0−cos10 | | cos10 | |
sin30*sin50 = |
| = |
| = |
| |
| | −2 | | −2 | | 2 | |
5sin100 = 5sin(90+10) = 5*[ sin90*cos10 + sin10*cos90 ] =
5*[ 1*cos10 + 0*sin10 ] = 5cos10
| sin30*cos50 | | | | cos10 | | 1 | |
| = |
| = |
| = |
| |
| 5sin100 | | 5cos10 | | 10cos10 | | 10 | |
5 mar 22:06
shersken: dzieki wielkie^^
5 mar 22:17
Agniesia: Basiu jeśli mogę się wtrącić.
x+y = 100
x − y = 60
x = 80
y = 20
moim zdaniem
5 mar 22:22