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Katyalee08
@Katyalee08
August 2022
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упростите tgB+1/1+ctgB
Докажите,что при всех допустимых значениях альфа выражение принимает одно и то же значение:
а)(sina+cosa)2-2sina*cosa
б) sin4a+cos4a+2sin2a*cosa
в) 2-sin2a-cos2a/3sin2a+3cos2a
г)sin4a-cos4a/sin2a-cos2a
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Answers & Comments
pomoshnik09
1) =(sinb/cosb+1)/(cosb/sinb+1)= (sinb+cosb)/cosb):(sinb+cos b)/sinb=
=(sinb+cosb)/cosb)*(sinb/(sinb+cosb)=sinb/cosb=tgb
2) =²a+2sinacosa+cos²a-2sina* cosa=sin²a+cos²a=1
3)=(sin²a+cos²a)²=1
4) =2-(sin²a+cos²a)/3(sin²a+cos²a)=2-1/3*1=1/3
5) = (sin²a-cos²a)(sin²a+cos²a)/sin²a-cos²a)=sin²a+cos²a=1
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Answers & Comments
=(sinb+cosb)/cosb)*(sinb/(sinb+cosb)=sinb/cosb=tgb
2) =²a+2sinacosa+cos²a-2sina* cosa=sin²a+cos²a=1
3)=(sin²a+cos²a)²=1
4) =2-(sin²a+cos²a)/3(sin²a+cos²a)=2-1/3*1=1/3
5) = (sin²a-cos²a)(sin²a+cos²a)/sin²a-cos²a)=sin²a+cos²a=1