1) sin40° + cos70° = cos50° + cos70° = 2cos(50° + 70°)/2 · cos(50° - 70°)/2 = 2cos60° · cos10°
Если упростить: sin40° + cos70° = cos50° + cos70° = 2cos(50° + 70°)/2 · cos(50° - 70°)/2 = 2cos60° · cos10° = 2·0,5· cos10° = cos10°
2) tg25° - ctg70° = tg25° - tg20° = sin(25° - 20°)/(cos25°·cos20°) = sin5°/(cos25°·cos20°)
Объяснение:
в 1)-м сos70=cos(90-20)=sin20 и тогда имеем sin40+sin20 и далее по формуле, ctg70=ctg(90-20)=tg20, tg25-tg20=
