Преобразуйте в произведение выражение: а)cos^2a-cos^2b - вопрос №2234221228941814 от дадададад2 04.03.2022 19:22

Question

Алгебра: Преобразуйте в произведение выражение: а)cos^2a-cos^2b б)3/4-sin^2x в)cos^2x-1/2 г)sin^2a-cos^2п/3 заранее )

дадададад2 · Answer

А)cos²α-cos²β=(cosα-cosβ)(cosα+cosβ)=-2sin(α-β)/2sin(α+β)/2·2cos(α+β)/2cos(α-β)/2=-2sin(α-β)/2cos(α-β)/2·2sin(α+β)/2cos(α+β)/2=-sin(α-β)sin(α+β)
б)3/4-sin²x=(√3/2)²-sin²x=(√3/2-sinx)(√3/2+sinx)=(sinπ/3-sinx)(sinπ/3+sinx)=2sin(π/3+x)/2cos(π/3-x)/2·2sin(π/3-x)/2cos(π/3+x)/2=2sin(π/3+x)/2cos(π/3+x)/2·2sin(π/3-x)/2cos(π/3-x)/2=sin(π/3+x)sin(π/3-x)
в)cos²x-1/2=cos²x-(√2/2)²=(cosx-√2/2)(cosx+√2/2)=(cosx-cosπ/4)(cosx+cosπ/4)=-2sin(x-π/4)/2sin(x+π/4)/2·2cos(x+π/4)/2cos(x-π/4)/2=-2sin(x-π/4)/2cos(x-π/4)/2·2sin(x+π/4)/2cos(x+π/4)/2=-sin(x-π/4)sin(x+π/4)
г)sin²α-cos²π/3=(sinα-cosπ/3)(sinα+cosπ/3)=(sinα-1/2)(sinα+1/2)=(sinα-sinπ/6)(sinα+sinπ/6)=2sin(α-π/6)/2cos(α+π/6)/2·2sin(α+π/6)/2cos(α-π/6)/2=2sin(α-π/6)/2cos(α+π/6)/2·2sin(α+π/6)/2cos(α+π/6)/2=sin(α-π/6)sin(α+π/6)