设三角形ABC的内角ABC的对边长分别为a、b、c,且3b的平方+3C的平方-3a的平方=4倍根号2bc①:求sinA的值②:求2sin(A+π/4)sin(B+C+π/4)/1-cos2A的值

问题描述:

设三角形ABC的内角ABC的对边长分别为a、b、c,且3b的平方+3C的平方-3a的平方=4倍根号2bc
①:求sinA的值
②:求2sin(A+π/4)sin(B+C+π/4)/1-cos2A的值

解析:∵3b^2+3c^2-3a^2=4√2bc,
∴(b^2+c^2-a^2)/2bc=2√2/3
即cosA=2√2/3
sinA=√[1-(cosA)^2]=√(1-8/9)=1/3
2sin(A+π/4)=√2(sinA+cosA)
=√2(1/3+2√2/3)=√2/3+4/3
sin(B+C+π/4)/(1-cos2A)
=sin(A-π/4)/(1-cos2A)
=√2/2(sinA-cosA)/2(sinA)^2
=√2/2(1/3-2√2/3)/(2/9)
=3√2/4-3