三角恒等公式的推导

1.sin^2(a)+sin^2(a+2pi/3)+sin^2(a-2pi/3)
=sin^2(a)+[1-cos(2a+4pi/3)]/2+[1-cos(2a-4pi/3)]/2
=sin^2(a)+1-[cos(2a+4pi/3)+cos(2a-4pi/3)]/2
=sin^2(a)+1-[cos2acos(4pi/3)-sin2asin(4pi/3)+cos2acos4pi/3+sin2asin(4pi/3)]/2
=[1-cos2a]/2 +1-cos2a*cos(4pi/3)
= -cos2a/2-cos2a*(-cospi/3)+3/2
=-cos2a/2+cos2a/2+3/2
=3/2
2.sin^3(a)+sin^3(a+2pi/3)+sin^3(a-2pi/3)
=sin^3(a)+[sin(a+2pi/3)+sin(a-2pi/3)]*[
sin^2(a+2pi/3)+sin^2(a-2pi/3)-sin(a+2pi/3)*sin(a-2pi/3)]
=sin^3(a)+[2sinacos(2pi/3)]*[3/2-sin^2(a)
-sin^2(a)cos^2(2pi/3)+cos^2(a)sin^2(2pi/3)]
=sin^3(a)+[-sina]*[3/2-5/4sin^2(a)+3/4cos^2(a)]
=sin^3(a)+[-sina]*[9/4-2sin^2(a)]
=sin^3(a)-9/4sina+2sin^3(a)
=-3/4*[-4sin^3(a)+3sina]
=-3/4*sin3a
注:sin3a=sin(2a+a)=sin2acosa+cos2asina=(2sinacosa)cosa+[1-2(sina)^2]sina=2sina[1-(sina)^2]+[1-(sina)^2]sina=3sina-4(sina)^3