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urgent 20分數學 三角恆等式

發問:

1. sin freeter=9over41 求 3cos freeter +2 over 3 tan freeter +3 的值2.sin(90-freeter)x cos(90-freeter) over tan freeter + cos^2(90-freeter)3.1-cos^2 36 over cos^2 544. sin(90-freeter) cos (90- freeter) tan freeter x+ sin^2 (90-freeter)5.1 over sin^2 34 - tan^2 266. (1+sin freeter)(1-sin freeter)7.tan^2 freeter... 顯示更多 1. sin freeter=9over41 求 3cos freeter +2 over 3 tan freeter +3 的值 2.sin(90-freeter)x cos(90-freeter) over tan freeter + cos^2(90-freeter) 3.1-cos^2 36 over cos^2 54 4. sin(90-freeter) cos (90- freeter) tan freeter x+ sin^2 (90-freeter) 5.1 over sin^2 34 - tan^2 26 6. (1+sin freeter)(1-sin freeter) 7.tan^2 freeter (1-sin^2 freeter) 8. sin freeter over 1+ cos freeter -1-cos freeter over sin freeter 9 tan49 - 1 over tan 41 (數字係度數黎既 ^2 係2次方

最佳解答:

1. sin θ=9/41 求 3cos θ +2/3 tan θ +3 的值 sin θ=9/41 cos θ = (√412 - 92)/41 cos θ = 40/41 ∴3cos θ +2/3 tan θ +3 =3(40/41) + 2/3 (9/40) + 3 =4983/820 2.sin(90-θ) x cos(90-θ) / tan θ + cos2(90-θ) =cos θ x sin θ / (sin θ/cos θ) + sin2θ =cos2θ +sin2θ =1 3.1-cos236 / cos254 =1-cos236 / cos254 =sin236/cos254 =cos254/cos254 =1 4. sin(90-θ) cos (90-θ) tan θ x+ sin^2 (90-θ) sin(90-θ) cos (90-θ) tan θ + sin^2 (90-θ)嗎 圖片參考：http://hk.yimg.com/i/icon/16/7.gif cos θ sin θ tan θ + cos2θ =cos θ sin θ sin θ/cos θ + cos2θ =sin2θ + cos2θ =1 5.1/ sin234 - tan226 1/ sin234 - tan256 圖片參考：http://hk.yimg.com/i/icon/16/7.gif =1/sin234 - 1/tan234 =1/sin234 - cos234/sin234 =(1-cos234)/sin234 =sin234/sin234 =1 6. (1+sin θ)(1-sin θ) (1+sin θ)(1-sin θ) =12 - sin2θ =cos2θ 7.tan2θ (1-sin2θ) tan2θ (1-sin2θ) =sin2θ/cos2θ x cos2θ =sin2θ 8. sin θ / 1+ cos θ -1-cos θ / sin θ sin θ / (1+ cos θ) (-1-cos θ) / sin θ =- sin θ / (1+ cos θ) x (1+cos θ) / sin θ =-1 9 tan49 - 1 / tan 41 tan49 - 1 / tan 41 =tan 49 - tan 49 =0

其他解答:

1. sinθ=9/41 求 3cosθ+2 / 3 tan θ +3 的值 cosθ = 40/41 AND tanθ = 9/40 3cosθ + 2 / 3tanθ + 3 = 3(40/41) + 2 / 3(9/40) + 3 = 202/41 / 147/40 = 8080/6027 2. [sin(90 - θ) x cos(90 - θ) / tan θ] + cos^2(90 - θ) = (cosθ x sinθ / tanθ) + sin^2(θ) = cos^2θ + sin^2θ = 1 3. 1-cos^2 36 / cos^2 54 = sin^2(36) / sin^2(90 -54) = sin^2(36) / sin^2(36) = 1 4. sin(90 - θ) cos (90 - θ) tan θ + sin^2 (90 - θ) = sin^2θ + cos^2θ = 1 5. 1 / sin^2 (34) - tan^2 (26) 題目不清 6. (1+sin θ)(1-sinθ) = 1 - cos^2θ = sin^2θ 7. tan^2 θ (1-sin^2 θ) = tan^2θ (cos^2θ) = sin^2θ 8. [sinθ / (1+ cos θ)] - (1- cosθ / sin θ) = [sin^2θ - (1 - cosθ)(1 + cosθ)] / sinθ(1 + cosθ) = sin^2θ - sin^2θ / sinθ(1 + cosθ) = 0 9 tan49 - 1/ tan 41 = tan(90 - 41) - 1/tan41 = 1/tan41 - 1/tan41 = 00D7DAC4E6DF60DFA