已知向量AB=(1+tanx,1-tanx),向量AC=(sin(x-π/4),sin(x+π/4)
问题描述:
已知向量AB=(1+tanx,1-tanx),向量AC=(sin(x-π/4),sin(x+π/4)
(1)求证:角BAC为直角
(2)若x属于[-π/4,π/4],求三角形ABC的边BC的长度的取值范围
答
1.证明:角BAC为直角,即,证明:向量AB*向量AC=0,即可,向量AB*向量AC=(1+tanx)*sin(x-π/4)+(1-tanx)*sin(x+π/4)=[sin(x-π/4)+sin(x+π/4)]+tanx[sin(x-π/4)-sin(x+π/4)]=2*sinx*cos(-π/4)+tanx*2cosx*sin(-π/4)=...