Question

直角三角形的充要条件

在三角形ABC中,求证三角形ABC为直角三角形的充要条件是sin2A+sin2B+sin2C=2
要一个精确的答案

Answer 1

首先,sin(180°-X）=sinX,cos（180°-X）=-cosX,sin(90°-X)=cosX,cos(90°-X)=sinX ①sin^2A sin^2B-sin^2C =sin^2A sin^2B-sin^2（180°-（A B）） =sin^2A sin^2B-sin^2（A B） =sin^2A sin^2B-(sinAcosB sinBcosA)^2 =sin^2A sin^2B-sin^2Acos^2B-sin^2Bcos^2A-2sinAsinBcosAcosB =sin^2A(1-cos^2B) sin^2B(1-cos^2A)-2sinAsinBcosAcosB =sin^2Asin^2B sin^2Bsin^2A-2sinAsinBcosAcosB =2sinAsinB(sinAsinB-cosAcosB) =2sinAsinB(-cos(A B)) =2sinAsinBcos(180°-（A B）) =2sinAsinBcosC ②sinA sinB-sinC =（sinA sinB）-sin(180°-（A B）) =（sinA sinB）-sin(A B) =2sin((A B)/2)cos((A-B)/2) -2sin((A B)/2) cos((A B)/2) =2sin((A B)/2) (cos((A-B)/2- cos((A B)/2)) =2cos(90°-(A B)/2)2（ -2sin(((A-B)/2-(A B)/2)/2) sin(((A-B)/2 (A B)/2)/2)) =2cos((180°-(A B))/2) (-2sin(-B/2)sin(A/2)) =4cos(C/2)sin(B/2)sin(A/2)