[ KitanovV @ 03.07.2019. 12:09 ] @
| Uprostiti izraz: (sin3+sin3x):sinx+(cos3x-cos3x):cosx |
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[ KitanovV @ 03.07.2019. 12:09 ] @
[ mjanjic @ 03.07.2019. 16:21 ] @
(sin3x+sin3x):sinx+(cos3x-cos3x):cosx
sin3x = 3sinx - 4sin3x cos3x = cos3x - 3cosxsin2x [ KitanovV @ 03.07.2019. 16:38 ] @
Jel rezultat 1.
[ mjanjic @ 03.07.2019. 20:03 ] @
3
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