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mirasimashev
@mirasimashev
September 2021
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ПОмогите решиить
[tex]cos4a-2cos2a+2- cos^{2}2a [/tex]
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kirichekov
Verified answer
Cos4α-2cos2α+2-cos²2α=
4sin
⁴α
1. cos4α=cos(2*2α)=cos²2α-sin²2α
2. -2cos2α+2=-2*(cos2α-1)=-2*(cos²α-sin²α-sin²α-cos²α)=4sin²α
3. cos²2α-sin²2α+4sin²α-cos²2α=-(2sinα*cosα)²+4sin²α=-4sin²α*cos²α+4sin²α=-4sin²α *(cos²α-1)=-4sin²α*(-sin²α)=4sin⁴α
0 votes
Thanks 1
mirasimashev
спасибо. 4sin^4a
kirichekov
спасибо, сейчас исправлю
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Answers & Comments
Verified answer
Cos4α-2cos2α+2-cos²2α=4sin⁴α1. cos4α=cos(2*2α)=cos²2α-sin²2α
2. -2cos2α+2=-2*(cos2α-1)=-2*(cos²α-sin²α-sin²α-cos²α)=4sin²α
3. cos²2α-sin²2α+4sin²α-cos²2α=-(2sinα*cosα)²+4sin²α=-4sin²α*cos²α+4sin²α=-4sin²α *(cos²α-1)=-4sin²α*(-sin²α)=4sin⁴α