| tgx | ||
sinx= | = (−3/4)/(5/4)= −3/5 | |
| p[1+tg2x |
| 3 | ||
ale sinus w 2 cwiartce jest dodatni wiec sinx= | = 0,6 | |
| 5 |
| 1 | 4 | |||
cosx= | = 1/(5/4)= | |||
| p[1+tg2x | 5 |
| 4 | ||
Ale cos w 2 cwaiartce jest ujemny wiec cos x= − | = −0,8 | |
| 5 |
| y | 3 | |||
tgα= | = | r=√32+42=5 | ||
| x | −4 |
| y | 3 | 9 | ||||
sinα= | = | to sin2α= | ||||
| r | 5 | 25 |
| 9 | 34 | |||
W= sin2α+ sin2α+cos2α= sin2α+1 = | +1= | |||
| 25 | 25 |
| 9 | ||
Poprawiam sin2α= | ||
| 25 |
| x | 4 | 16 | ||||
cosα= | = − | to cos2α= | ||||
| r | 5 | 25 |
| 9 | 16 | |||
W= 2* | * | =........ | ||
| 25 | 25 |