ICSP: Ponieważ:
(sinx + cosx)
2 = sin
2x + cos
2x + 2sinxcosx = 1 + 2sinxcosx
więc
| (sinx + cosx)2 − 1 | |
a) sinx*cosx = |
| = ... |
| 2 | |
b) sin
4x + cos
4x = (sin
2x + cos
2x)
2 − 2(sinx*cosx)
2 = 1 − 2(sinx*cosx)
2 = ...
c)sin
3x + cos
3x = (sinx + cosx)(sin
2x − sinxcosx + cos
2x) = (sinx+cosx)(1 − sinxcosx) = ...
d) |sinx − cosx| =
√(sinx − cosx)2 =
√1 − 2sinxcosx