Question

已知sinc+cosc=2sina,sinc*cosc=sin^b,求证:4cos^2 2a=cos^2 2b

Answer 1

证明：
∵在三角形abc中，
∴a+b+c=180度，得sina=sin（b+c）
则a/2=90度-（b+c）/2,得cosa/2=sin（（b+c）/2）
左边=sin(b+c)+sinb+sinc
则4cos(a/2)cos(b/2)cos(c/2)
=4sin((b+c)/2)cos(b/2)cos(c/2)
=4cos(b/2)cos(c/2)(sinb/2·cosc/2+cosb/2·sinc/2)
=4sin(b/2)cos(b/2)(cos(c/2))^2+4sin(c/2)cos(c/2)(cos(b/2))^2
=sinb(cosc+1)+sinc(cosb+1)
=sin(b+c)+sinb+sinc
左边=右边
原式成立！