irena_1:
sinα+cosα=1,25
(sinα+cosα)
2=1,5625
sin
2α+cos
2α+2sinαcosα=1,5625
1+2sinαcosα=1,5625
2sinαcosα=0,5625
a)
(sinα−cosα)
2=sin
2α+cos
2α−2sinαcosα=1−0,5625=0,4325
b)
| | 1 | | 1 | |
sin4α+cos4α=(sin2α+cos2α)2−2sin2αcos2α=12− |
| *(2sinαcosα)2=1− |
| *0,56252= |
| | 2 | | 2 | |
| | 1 | | 9 | | 81 | | 81 | | 431 | |
=1− |
| *( |
| )2=1− |
| =1− |
| = |
| |
| | 2 | | 16 | | 2*256 | | 512 | | 512 | |