真多啊
1.用角ABC的正弦,余弦表示．
(1)Sin(A+B+C)
=sin[A+(B+C)]
=sinAcos(B+C)+cosA*sin(B+C)
=sinA(cosBcosC-sinBsinC)+cosA*(sinBcosC+cosBsinC)
=sinAcosBcosC-sinAsinBsinC+cosAsinBcosC+cosAcosBsinC
(2)COS(A+B+C)
=cos[A+(B+C)]
=cosAcos(B+C)-sinAsin(B+C)
=cosA(cosBcosC-sinBsinC)-sinA(sinBcosC+cosBsinC)
=cosAcosBcosC-cosAsinBsinC-sinAsinBcosC-sinAcosBsinC
2.用角ABC的正切表示tg(A+B+C)
tg(A+B+C)
=tg[A+(B+C)]
=[tgA+tg(B+C)]/[1-tgA*tg(B+C)]
=tgA+[(tgB+tgC)/(1-tgBtgC)]/[1-tgA*(tgB+tgC)/(1-tgBtgC)]
=[tgA(1-tgBtgC)+tgB+tgC]/[1-tgBtgC-tgA*(tgB+tgC)]
=(tgA+tgB+tgC-tgAtgBtgC)/(1-tgAtgB-tgBtgC-tgAtgC)
3.证明
(1)cos3a=4cos'3a-3cosa('3为指数）
cos3a
=cos(2a+a)
=cos2a*cosa-sin2asina
=(2cosa^2-1)cosa-2sinacosasina
=(2cosa^2-1)cosa-2(sina^2)cosa
=(2cosa^2-1)cosa-2(1-cosa^2)cosa
=4cosa^3-3cosa
实际上可以直接利用第1小题的证明
cos3a
=cos(a+a+a)
=cosa*cosa*cosa-cosa*sina*sina-sinasinacosa-sinacosasina
=4cosa^3-3cosa 注意：sina^2=1-cosa^2
(2)tan3a=(3tana-tan'3a)/(1-3tan'2a)('3,'2为指数)
tg3a
=tg(a+a+a) 直接利用第2题的公式
=(tga+tga+tga-tga*tga*tga)/(1-tga*tga-tga*tga-tga*tga)
=(3tga-tga^3)/(1-3tga^2)