用角A,B,C的正弦,余弦,正切表示:sin(A+B+C);cos(A+B+C);tan(A+B+C)
用角A,B,C的正弦,余弦,正切表示:sin(A+B+C);cos(A+B+C);tan(A+B+C)
最佳回答
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
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)
(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
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)
最新回答共有2条回答
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2026-04-02 20:03:07留胡子的白开水
回复用角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 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)
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