Задача 62096 ...

6035fc7016e9eb62f3f77539

11.12.2021 17:35:30

Докажите тождество:

(cos^2 2α -4cos^2 α +3)/(cos^2 2 α +4cos^2 α -1)=tg^4 α

5f3ffd6fb5ac0b3cc606fb26

11.12.2021 21:18:48

★

(cos^2(2a)+4cos^2(a)+3)/(cos^2(2a)+4cos^2(a)-1))=
=((1-2sin^2(a))^2-4(1-sin^2(a)+3)/((2cos^2(a)-1)^2+4cos^2(a)-1)=
=(1-4sin^2(a)+4sin^4(a)-4+4sin^2(a)+3)/(4cos^4(a)-4cos^2(a)+1+4cos^2(a)-1)=
=4sin^4(a)/4cos^4(a)=tg^4(a).
tg^4(a)=tg^4(a). Тождество доказано.РS ( cos2a=1-sin^2(a); cos^2(a)=1-sin^2(a) ).