三角函数的化简求值证明21.已知sin(a+B)cosa-1/2[sin(2a+B)-cosB]=1/2,0

1.
sin(2a+B)-cosB=sin[(a+B)+a]-cos[(a+B)-a]
=sin(a+B)cosa+cos(a+B)sina-cosB
所以:
sin(a+B)cosa-1/2[sin(2a+B)-cosB]
=sin(a+B)cosa-1/2sin(a+B)cosa-1/2cos(a+B)sina+1/2cosB
=1/2[sin(a+B)cosa-cos(a+B)sina]+1/2cosB
=1/2(sin(a+B-a)+cosB)
=1/2(sinB+cosB)
=1/2
就是有:sinB+cosB=1
平方一下,就得到sinBcosB=0
又0

令a+B 为x 则B为x-a
sinxcosa-1/2[sin(x+a)-cosB]=1/2
2sinxcosa-sin(x+a)+cosB=1
sinxcosa-sinacosx+cosB=1
sin(x-a)+cosB=1
sinB+cosB=1
所以 B=π/2