| 1 | 1 | 1 | √3 | |||||
sin(4 | π)=sin(4π+ | π)=sin( | π)= | |||||
| 3 | 3 | 3 | 2 |
| 3 | 3 | 3 | 3 | |||||
ctg(−1 | π)=ctg(−π− | π)=ctg(− | π)=−ctg( | π)=−*(−1)=1 | ||||
| 4 | 4 | 4 | 4 |
| π | π | √3 | ||||
sin(4π+ | )=sin | = | (skorzystałam z okresowości funkcji sinus) | |||
| 3 | 3 | 2 |
| 3 | 3 | π | π | |||||
ctg(−134π)=−ctg(π+ | π)=−ctg | π=−ctg(π−π4)=−(−ctg | )=ctg | =1 | ||||
| 4 | 4 | 4 | 4 |