| 5 | ||
kąt α jest ostry i cos α= | .Oblicz tg2+1 | |
| 6 |
| sin x | sin2 x | 1− cos2 x | ||||
lub tg2 x + = ( | )2 +1 = | +1 = | + 1 = | |||
| cos x | cos2 x | cos2x |
| 1− cos2 x | cos2 x | 1 | ||||
= | + | = | ||||
| cos2x | cos2 x | cos2 x |
| sin2α | 1136 | |||
tg2α+1= | +1= | +1=1125+1=.... | ||
| cos2α | 2536 |
| 5 | ||
Zamiast cosα wpisujesz | , a cos2α to jest to samo co (cosα)2 | |
| 6 |
| 5 | ||
sin2α+( | )2=1 | |
| 6 |
| 25 | ||
sin2α=1− | ||
| 36 |
| 11 | ||
sin2α= | ||
| 36 |
| 5 | ||
Skoro cosα= | to | |
| 6 |
| 5 | 25 | |||
cos2α=( | )2= | . | ||
| 6 | 36 |
| sinα | sin2α | |||
Jest wzór tgα= | , a tg2α= | |||
| cosα | cos2α |