Вот вторая часть.Помогите.. Created by Fredyson. algebra-ru

Вот вторая часть.Помогите...

Fredyson @Fredyson

October 2021 1 16 Report

Вот вторая часть.Помогите.

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sedinalana

Verified answer

1
-sin²a/cos²a -1=-tg²a-1=-(tg²a+1)=-1/cos²a
2
2cos(π/3+a)-2(1/2cosa-√3/2sina)=2cos(π/3+a)-2(cosπ/3cosa-sinπ/3sina)=
=2cos(π/3+a)-2cos(π/3+a)=0
3
2sin2acos4a/2cos2acos4a=tg2a

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Answers & Comments

Verified answer

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