Loading [MathJax]/jax/output/CommonHTML/jax.js
Đăng ký

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=4sinsinβsinγ

d) cos2+cos2β+cos2γ=12coscosβcosγ

 

Hướng dẫn giải

a) Ta có:

sinα+sinβ+sinγ=sinα+2sinβ+γ2cosβγ2=sinα+2sinπα2cosβγ2=2sinα2cosα2+2cosα2cosβγ2=2cosα2(sinα2+cosβγ2)=2cosα2[sinπ(β+γ)2+cosβγ2]=2cosα2(cosβ+γ2+cosβγ2)=4cosα2cosβ2cosγ2

b) Ta có:

cosα+cosβ+cosγ=2cosα+β2cosαβ2+12sin2γ2=2cos(π2γ2)cosαβ2+12sin2γ2=1+2sinγ2(cosαβ2sinγ2)=1+2sinγ2(cosαβ2cosα+β2)=1+4sinα2sinβ2sinγ2

c) sin2+sin2β+sin2γ

=2sin(+β)cos(β)+2sinγcosγ

=2sinγ(cos(β)cos(+β))

=4sinsinβsinγ

d) Ta có:

cos2+cos2β+cos2γ=1+cos2α2+1cos2β2+cos2γ=1+12(cos2α+cos2β)+cos2γ=1+cos(α+β)cos(αβ)+cos2γ=1+cosγ(cosγcos(αβ))=1cosγ[cos(α+β)+cos(αβ)]=12coscosβcosγ