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catkama
@catkama
July 2022
1
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Помогите решить пожалуйста)Даю 25 баллов))
sin^2(35)-cos^2(145)/cos100*cos350
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oganesbagoyan
Verified answer
Task/28346589
-------------------
( sin
²35°
- cos
²145°
) / cos100
°*cos350° =
( (1- cos2*35°) / 2- (1+cos2*145°) / 2 ) / cos(90°+10°)*cos(360°-10°) =
- ( cos290°+cos70°) / -2 sin10°*cos10°=( cos(360°-70°) + cos70°) / 2sin20° =
(cos70°+cos70°)/2sin20°=cos70°/sin20°=sin20°/sin20°
=1 .
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Verified answer
Task/28346589-------------------
( sin²35° - cos²145°) / cos100°*cos350° =
( (1- cos2*35°) / 2- (1+cos2*145°) / 2 ) / cos(90°+10°)*cos(360°-10°) =
- ( cos290°+cos70°) / -2 sin10°*cos10°=( cos(360°-70°) + cos70°) / 2sin20° =
(cos70°+cos70°)/2sin20°=cos70°/sin20°=sin20°/sin20° =1 .