Задача 55843 ...

5fc0a63d64081d1dafcb2733

27.11.2020 10:10:31

Упростить выражение log₂(((sinα - cosα)² - 1 + ½ sin2α)² + ¼ cos²2α)

5f3ea7e3faf909182968ddd9

27.11.2020 13:23:15

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(sin α -cos α )^2=sin^2 α -[blue]2sin α *cos α [/blue]+cos^2 α =(sin^2 α+cos^2 α)-[blue]sin2 α[/blue] =1-sin2 α

(sin α -cos α )^2-1+(1/2)sin2 α =1-sin2 α -1+(1/2)sin2 α =-(1/2)sin2 α

[b]([/b](sin α -cos α )^2-1+(1/2)sin2 α[b])[/b]^2=[b]([/b](-1/2)sin2 α [b])[/b]^2=(1/4)sin^22 α


[b]([/b](sin α -cos α )^2-1+(1/2)sin2 α[b])[/b]^2+(1/4)cos^22 α =(1/4)sin^22 α +(1/4)sin^22 α=(1/4)*(sin^2 2α+cos^2 2α)=(1/4)*1=(1/4)


log_(2)([b]([/b](sin α -cos α )^2-1+(1/2)sin2 α[b])[/b]^2+(1/4)cos^22 α )=log_(2)(1/4)=[red][b]-2[/b][/red]- о т в е т