Bài 51 trang 216 SGK Đại số 10 Nâng cao
Đề bài
Chứng minh rằng nếu ∝+β+γ=π thì
a) sinα+sinβ+sinγ=4cosα2cosβ2cosγ2
b) cosα+cosβ+cosγ=1+4sinα2sinβ2sinγ2
c) sin2∝+sin2β+sin2γ=4sin∝sinβsinγ
d) cos2∝+cos2β+cos2γ=1–2cos∝cosβcosγ
Hướng dẫn giải
a) Ta có:
sinα+sinβ+sinγ=sinα+2sinβ+γ2cosβ−γ2=sinα+2sinπ−α2cosβ−γ2=2sinα2cosα2+2cosα2cosβ−γ2=2cosα2(sinα2+cosβ−γ2)=2cosα2[sinπ−(β+γ)2+cosβ−γ2]=2cosα2(cosβ+γ2+cosβ−γ2)=4cosα2cosβ2cosγ2
b) Ta có:
cosα+cosβ+cosγ=2cosα+β2cosα−β2+1−2sin2γ2=2cos(π2−γ2)cosα−β2+1−2sin2γ2=1+2sinγ2(cosα−β2−sinγ2)=1+2sinγ2(cosα−β2−cosα+β2)=1+4sinα2sinβ2sinγ2
c) sin2∝+sin2β+sin2γ
=2sin(∝+β)cos(∝−β)+2sinγcosγ
=2sinγ(cos(∝−β)−cos(∝+β))
=4sin∝sinβsinγ
d) Ta có:
cos2∝+cos2β+cos2γ=1+cos2α2+1cos2β2+cos2γ=1+12(cos2α+cos2β)+cos2γ=1+cos(α+β)cos(α−β)+cos2γ=1+cosγ(cosγ−cos(α−β))=1−cosγ[cos(α+β)+cos(α−β)]=1−2cos∝cosβcosγ