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kruss802
@kruss802
July 2022
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Vasily1975
Verified answer
№ 9.54
а) =cos²π/8-sin²π/8=cosπ/4=√2/2
б) =cos²π/12-sin²π/12=cosπ/6=√3/2
№ 9.55
a)=(sin2α*cos2α)/2=1/4*sin4α
б)=(cos²α+sin²α)(cos²α-sin²α)=cos²α-sin²α=cos2α
в)=(sin²α+2sinα*cosα+cos²α)/(1+sin2α)=(1+sin2α)/(1+sin2α)=1
г) =(cos²α-sin²α)/(sinα-cosα)=(cosα+sinα)(cosα-sinα)/(sinα-cosα)=-(cosα+sinα)=-cosα-sinα
д) = 2cos²α-(cos²α-sin²α)=cos²α+sin²α=1
е) = cos²α-sin²α+2sin²α=cos²α+sin²α=1
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Answers & Comments
Verified answer
№ 9.54а) =cos²π/8-sin²π/8=cosπ/4=√2/2
б) =cos²π/12-sin²π/12=cosπ/6=√3/2
№ 9.55
a)=(sin2α*cos2α)/2=1/4*sin4α
б)=(cos²α+sin²α)(cos²α-sin²α)=cos²α-sin²α=cos2α
в)=(sin²α+2sinα*cosα+cos²α)/(1+sin2α)=(1+sin2α)/(1+sin2α)=1
г) =(cos²α-sin²α)/(sinα-cosα)=(cosα+sinα)(cosα-sinα)/(sinα-cosα)=-(cosα+sinα)=-cosα-sinα
д) = 2cos²α-(cos²α-sin²α)=cos²α+sin²α=1
е) = cos²α-sin²α+2sin²α=cos²α+sin²α=1