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Пряник8Б
@Пряник8Б
August 2022
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Помогите пожалуйста(((
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pomoshnik09
2) 1/2*(2sinπ/12cosπ/12)=1/2*sin2π/12=1/2sinπ/6=1/2*1/2=1/4
3) 2*5sin32/2*cos16*cos(90-74)= 10sin32/2*cos16*sin16=10sin32/sin32=10
4)27cos2a=27*(cjs²a-sin²a)
по формуле sin²a+cos²a=1 найдем sin²a
(1/3)²+sin²a=1 sin²a=1-1/9=8/9
27*(cos²a-sin²a)=27*(1/9-8/9)=27*-7/9=-21
5) =-5(sin(2π-a)+2cos(3π/2+a)=5sina-2sina=3sina=3*(-0.1)=-0.3
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Answers & Comments
3) 2*5sin32/2*cos16*cos(90-74)= 10sin32/2*cos16*sin16=10sin32/sin32=10
4)27cos2a=27*(cjs²a-sin²a)
по формуле sin²a+cos²a=1 найдем sin²a
(1/3)²+sin²a=1 sin²a=1-1/9=8/9
27*(cos²a-sin²a)=27*(1/9-8/9)=27*-7/9=-21
5) =-5(sin(2π-a)+2cos(3π/2+a)=5sina-2sina=3sina=3*(-0.1)=-0.3