Question

问两道高一数学题，三角函数

1、已知π/4<α<3π/4，0<β<π/4，cos(π/4+α)=-3/5，sin(3π/4+β)=5/13，求sin(α+β)的值2、在三角形ABC中，（1）若cosB=3/5,cosC=5/13，求sinA（2）若sinA/2+sin(B/2+C/2)=√2，判断三角形的形状，说明理由

Answer 1

1.sin(π/4+α)=±√1-cos²(π/4+α) =±4/5
∵π/4<α<3π/4
∴π/2<π/4+α<π
∴sin(π/4+α)=4/5
同理：cos(3π/4+β)=-12/13
∵sin[(π/4+α)+(3π/4+β)]=sin[π+(α+β)]=-sin(α+β)
∴sin[(π/4+α)+(3π/4+β)]=sin(π/4+α)·cos(3π/4+β) + cos(π/4+α)·sin(3π/4+β)=-63/65
∴sin(α+β)=63/65

2.(1)sinB=±√1-cos²B=±4/5 ， sinC=±√1-cos²C=±12/13
∵是在三角形ABC中
∴sinB=4/5 ， sinC=12/13
sinA=sin[π-(B+C)]=sin(B+C)=sinBcosC + sinCcosB =56/65
(2)sin(A/2) + sin(B/2+C/2)=sin(A/2) + sin(π/2 - A/2) =sin(A/2) + cos(A/2)=√2
两边平方：sin²(A/2) + 2×sin(A/2)×cos(A/2) + cos²(A/2)=2
1 + sinA = 2
sinA=1
∴在三角形ABC中，∠A=90°
∴三角形ABC是直角三角形

Answer 2

①sin(π/4+α)=4/5,cos(3π/4+β)=-12/13
sin(α+β)=-sin(α+β+π)=-(sin(π/4)cos(3π/4+β)+cos(π/4)sin(3π/4+β)
=-(-48/65-15/65)=63/65
②⑴cos(B+C)=-cosA=cosBcosC-sinBsinC=-33/65
cosA=33/65∴sinA=56/65
⑵∵sin(B/2+C/2)=cos[π/2-(A+B)/2]=cosA/2
∴sinA/2+sin(B/2+C/2)=sinA/2+cosA/2=√2
∴1+2sinA/2cosA/2=2
∴2sinA/2cosA/2=sinA=1
∵在三角形中
∴A=π/2△ABC为RT△

Answer 3

1.∵<α<3π/4，0<β<π/4
∴π／2＜π/4+α＜π 3π／4＜3π/4+β＜π
π/4＜ α+β＜π
又∵cos(π/4+α))=-3/5， sin(3π/4+β)=5/13
∴sin(π/4+α)＝4／5 cos (3π/4+β)=﹣12／13
∴sin(α+β)= ﹣sin(π+α+β)=﹣sin[(π/4+α)+(3π/4+β)]
=﹣[sin(π/4+α)cos (3π/4+β)+cos(π/4+α)sin(3π/4+β)]
=-[(4／5)×(﹣12／13)＋(-3/5)×5/13]=63／65

Answer 4

A=45 直角三角形