| cosx | π | |||
tgx+ | +2=0 sinx≠−1 x≠ | +2kπ | ||
| 1+sinx | 2 |
| sinx+sin2x+cos2x | |
+2=0 | |
| cosx+sinxcosx |
| sinx+1 | |
+2=0 | |
| cosx+sinxcosx |
| π | ||
2sinx+2cosx=tgx+1 x≠ | +kπ k∊C | |
| 2 |
| sinx | ||
2sinx− | +2cosx−1=0 | |
| cosx |
| 1 | ||
Kąty α i β są kątami ostrymi. Spełniony jest warunek sin2α − | cosβ = 0. | |
| 3 |
| dx | ||
g) ∫ | ||
| 1+sin2 x |
| dx | ||
, h) ∫ | ||
| 5−4sin x +3cosx |
| dx | dx | (1−cos 2x)2 dx | ||||
d) ∫ | , e) ∫ | , f) ∫ | , | |||
| 1+3cos x | sinx (cosx)3 | (1+cos 2x)3 |
| dx | dx | |||
g) ∫ | , h) ∫ | |||
| 1+(sin x)2 | 5−4sin x +cosx |
| (m+2) | ||
Niech f(x) = | x3 − 4x2 + (8−m)x +5 i za zadanie mam określić: | |
| 3 |
| 1 | 1 | 1 | ||||
( | −sina)( | −cosa)( | +tga) | |||
| sina | cosa | tga |
| cosa+sina)(cosa−sina) | ||
Uzasadnij, że jeżeli a jest kątem ostrym to | = 1−tg2a | |
| cos2a |
| 1 | ||
ab + bc + ac < | ||
| 3 |
| 1 | ||
ab + bc + ac ≥ − | ||
| 3 |
| 2 | ||
ab + bc + ac ≤ | ||
| 3 |
| 1 | ||
ab + bc + ac ≤ | ||
| 3 |
| 4*1020−8*1010+4 | ||
(66...6)2 zamieniłem na takie coś | ||
| 9 |