证明:tan3/2X-tan1/2x=2sinx/(cosx+cos2x).我会证明,只是我只能找到一种解法,哪位哥哥姐
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证明:tan3/2X-tan1/2x=2sinx/(cosx+cos2x).我会证明,只是我只能找到一种解法,哪位哥哥姐姐能帮帮我
证明:tan(3/2)X-tan(1/2)x=2sinx/(cosx+cos2x).
证明一:左边={sin[(3/2)x]/cos(3/2)x}-{sin[(1/2)x]/cos(1/2)x}
={sin[(3/2)xcos(1/2)x-cos[(3/2)x]sin(1/2)x}/[cos(3x/2)cos(x/2)
=[sin(3x/2-x/2)]/{(1/2)[cos(3x/2+x/2)+cos(3x/2-x/2)]}
=2(sinx)/(cos2x+cosx)=右边
证明二:右边=[sin(3x/2-x/2)]/{(1/2)[cos(3x/2+x/2)+cos(3x/2-x/2)]}
={sin[(3/2)xcos(1/2)x-cos[(3/2)x]sin(1/2)x}/[cos(3x/2)cos(x/2)
={sin[(3/2)x]/cos(3/2)x}-{sin[(1/2)x]/cos(1/2)x}
=tan(3x/2)-tan(x/2)=左边.
证明一:左边={sin[(3/2)x]/cos(3/2)x}-{sin[(1/2)x]/cos(1/2)x}
={sin[(3/2)xcos(1/2)x-cos[(3/2)x]sin(1/2)x}/[cos(3x/2)cos(x/2)
=[sin(3x/2-x/2)]/{(1/2)[cos(3x/2+x/2)+cos(3x/2-x/2)]}
=2(sinx)/(cos2x+cosx)=右边
证明二:右边=[sin(3x/2-x/2)]/{(1/2)[cos(3x/2+x/2)+cos(3x/2-x/2)]}
={sin[(3/2)xcos(1/2)x-cos[(3/2)x]sin(1/2)x}/[cos(3x/2)cos(x/2)
={sin[(3/2)x]/cos(3/2)x}-{sin[(1/2)x]/cos(1/2)x}
=tan(3x/2)-tan(x/2)=左边.
证明:tan3/2X-tan1/2x=2sinx/(cosx+cos2x).我会证明,只是我只能找到一种解法,哪位哥哥姐
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