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數學問題一問..(thx)萬分感激!!!!!
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最佳解答:
(1) sin 20 + sin 40 = 2 sin [(20 + 40)/2] cos [(20 - 40)/2] = 2 sin 30 cos 10 = cos 10 = sin 80 sin 20 + sin 40 - sin 80 = 0 (2) (sin 20 - sin 40)/(cos 20 - cos 40) = 2 sin [(20 - 40)/2] cos [(20 + 40)/2]/{-2 sin [(20 + 40)/2] sin [(20 - 40)/2]} = -2 sin 10 cos 30/(2sin 30 sin 10) = - 1/tan 30 = -√3 (4) (cos 48 + cos 12)/(sin 48 + sin 12) = 2 cos 30 cos 18/(2 sin 30 cos 18) = 1/tan 30 = √3 (5) cos 40 cos 80 + cos 80 cos 160 + cos 160 cos 40 = cos 80(cos 40 + cos 160) + cos 160 cos 40 = 2 cos 80 cos 100 cos 60 + cos 160 cos 40 = cos 80 cos 100 + cos 160 cos 40 = (cos 180 + cos 20)/2 + (cos 200 + cos 120)/2 = -1/2 + cos 20/2 - cos 20/2 - 1/4 = -3/4 (6) cos2 73 + cos2 47 + cos 73 cos 47 = (cos 73 + cos 47)2 - cos 73 cos 47 = (2 cos 60 cos 13)2 - (cos 120 + cos 26)/2 = cos2 13 + 1/4 - cos 26/2 = cos2 13 + 1/4 - (2cos2 13 - 1)/2 = 3/4 (7) sin 20 cos 40 cos 80 cos 160 = - sin 20 cos 20 cos 40 cos 80 = -(1/2) sin 40 cos 40 cos 80 = -(1/4) sin 80 cos 80 = -(1/8) sin 160 = - sin 20/8 cos 40 cos 80 cos 160 = 1/8 (8) sin 20 cos 70 + sin 10 sin 50 = (sin 90 - sin 50)/2 + (cos 40 - cos 60)/2 = 1 - sin 50/2 + sin 50/2 - 1/4 = 3/4 (9) cos 48 + cos 24 - cos 12 - cos 84 = 2 cos 36 cos 12 - 2 cos 36 cos 48 = 2 cos 36 (cos 12 - cos 48) = 4 cos 36 sin 30 sin 18 = 2 cos 36 sin 18 = 2 cos 36 sin 18 cos 18/cos 18 = sin 36 cos 36/cos 18 = sin 72/(2 cos 18) = cos 18/(2cos 18) = 1/2 (10) sin A + sin 3A + sin 5A + sin 7A = (sin A + sin 7A) + (sin 3A + sin 5A) = 2 sin 4A cos 3A + 2 sin 4A cos A = 2 sin 4A (cos 3A + cos A) cos A + cos 3A + cos 5A + cos 7A = (cos A + cos 7A) + (cos 3A + cos 5A) = 2 cos 4A cos 3A + 2 cos 4A cos A = 2 cos 4A (cos 3A + cos A) (sin A + sin 3A + sin 5A + sin 7A)/(cos A + cos 3A + cos 5A + cos 7A) = tan 4A 2010-02-06 22:18:20 補充： (3) cos10cos30cos50cos70=sin20sin40sin60sin80 考慮: sin20sin40sin80=(1/2)sin40(cos60-cos100) =(1/4)sin40-(1/2)sin40cos100 =(1/4)sin40-(1/4)(sin140-sin60) =(1/4)sin40-(1/4)sin40+(1/4)sin60 =(√3)/8 所以: cos10cos30cos50cos70=sin20sin40sin60sin80 =(√3)/8 x (√3)/2 =3/16
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數學問題一問..(thx)萬分感激!!!!!
發問:
此文章來自奇摩知識+如有不便請留言告知
和差化積與積化和差:1).求值:sin20`+sin40`-sin80`=_____.2).求值:(sin20`-sin40`)/(cos20`-cos40`)=_____.3).求值:cos10`cos30`cos50`cos70`=_____.4).求值:(cos48`+cos12`)/(sin48`+sin12`)=_____.5).求值:cos40`cos80`+cos80`cos160`+cos160`cos40`=_____.6).求值:cos^2 73`+cos^2 47`... 顯示更多 和差化積與積化和差: 1).求值:sin20`+sin40`-sin80`=_____. 2).求值:(sin20`-sin40`)/(cos20`-cos40`)=_____. 3).求值:cos10`cos30`cos50`cos70`=_____. 4).求值:(cos48`+cos12`)/(sin48`+sin12`)=_____. 5).求值:cos40`cos80`+cos80`cos160`+cos160`cos40`=_____. 6).求值:cos^2 73`+cos^2 47` +cos73`cos47`=_____. 7).求值:sin40`sin80`sin160`=_____. 8).求值:sin20`cos70`+sin10`sin50`=_____. 9).求值:cos48`+cos24`-cos12`-cos84`=_____. 10).證明:(sinA+sin3A+sin5A+sin7A)/(cosA+cos3A+cos5A+cos7A)=tan4A最佳解答:
(1) sin 20 + sin 40 = 2 sin [(20 + 40)/2] cos [(20 - 40)/2] = 2 sin 30 cos 10 = cos 10 = sin 80 sin 20 + sin 40 - sin 80 = 0 (2) (sin 20 - sin 40)/(cos 20 - cos 40) = 2 sin [(20 - 40)/2] cos [(20 + 40)/2]/{-2 sin [(20 + 40)/2] sin [(20 - 40)/2]} = -2 sin 10 cos 30/(2sin 30 sin 10) = - 1/tan 30 = -√3 (4) (cos 48 + cos 12)/(sin 48 + sin 12) = 2 cos 30 cos 18/(2 sin 30 cos 18) = 1/tan 30 = √3 (5) cos 40 cos 80 + cos 80 cos 160 + cos 160 cos 40 = cos 80(cos 40 + cos 160) + cos 160 cos 40 = 2 cos 80 cos 100 cos 60 + cos 160 cos 40 = cos 80 cos 100 + cos 160 cos 40 = (cos 180 + cos 20)/2 + (cos 200 + cos 120)/2 = -1/2 + cos 20/2 - cos 20/2 - 1/4 = -3/4 (6) cos2 73 + cos2 47 + cos 73 cos 47 = (cos 73 + cos 47)2 - cos 73 cos 47 = (2 cos 60 cos 13)2 - (cos 120 + cos 26)/2 = cos2 13 + 1/4 - cos 26/2 = cos2 13 + 1/4 - (2cos2 13 - 1)/2 = 3/4 (7) sin 20 cos 40 cos 80 cos 160 = - sin 20 cos 20 cos 40 cos 80 = -(1/2) sin 40 cos 40 cos 80 = -(1/4) sin 80 cos 80 = -(1/8) sin 160 = - sin 20/8 cos 40 cos 80 cos 160 = 1/8 (8) sin 20 cos 70 + sin 10 sin 50 = (sin 90 - sin 50)/2 + (cos 40 - cos 60)/2 = 1 - sin 50/2 + sin 50/2 - 1/4 = 3/4 (9) cos 48 + cos 24 - cos 12 - cos 84 = 2 cos 36 cos 12 - 2 cos 36 cos 48 = 2 cos 36 (cos 12 - cos 48) = 4 cos 36 sin 30 sin 18 = 2 cos 36 sin 18 = 2 cos 36 sin 18 cos 18/cos 18 = sin 36 cos 36/cos 18 = sin 72/(2 cos 18) = cos 18/(2cos 18) = 1/2 (10) sin A + sin 3A + sin 5A + sin 7A = (sin A + sin 7A) + (sin 3A + sin 5A) = 2 sin 4A cos 3A + 2 sin 4A cos A = 2 sin 4A (cos 3A + cos A) cos A + cos 3A + cos 5A + cos 7A = (cos A + cos 7A) + (cos 3A + cos 5A) = 2 cos 4A cos 3A + 2 cos 4A cos A = 2 cos 4A (cos 3A + cos A) (sin A + sin 3A + sin 5A + sin 7A)/(cos A + cos 3A + cos 5A + cos 7A) = tan 4A 2010-02-06 22:18:20 補充： (3) cos10cos30cos50cos70=sin20sin40sin60sin80 考慮: sin20sin40sin80=(1/2)sin40(cos60-cos100) =(1/4)sin40-(1/2)sin40cos100 =(1/4)sin40-(1/4)(sin140-sin60) =(1/4)sin40-(1/4)sin40+(1/4)sin60 =(√3)/8 所以: cos10cos30cos50cos70=sin20sin40sin60sin80 =(√3)/8 x (√3)/2 =3/16
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