jak to obliczyc?!
Iza19: sin20cos70 + cos20sin70 − tg10tg80 =
26 mar 17:56
Beti: sin20 = sin(90−70) = cos70
cos20 = j/w = sin70
tg10 = j/w = ctg80
czyli: cos270 + sin270 − ctg80tg80 = 1 − 1 = 0
26 mar 17:59
Święty: sin20=cos(90−20)=cos70
cos20=sin(90−20)=sin70
tg10=ctg(90−10)=ctg80
cos70+sin70−ctg80*tg80=1−1=0
26 mar 17:59
Johny Cape: sinαcosβ+sinβcosα= sin(α+β), więc sin20cos70+cos20sin70=sin90=1
tg80=ctg10, bo tgα=ctg(90−α)
tg10ctg10=tg10*1/tg10=1, (ctg=1/tg)
więc sin20cos70+cos20sin70 − tg10tg80= 1−1=0 jeśli dobrze rozumuję
26 mar 18:04
26 mar 19:09