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ruslank1460

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°)

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table123

Ответ:

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=

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