請計
2cos^2 45°-tan45° /sin^2 50°
急.......計sin cos tan
2009-05-04 4:44 am
回答 (2)
2009-05-04 5:17 am
✔ 最佳答案
2cos^2 45=1tan45 =1
0/sin^2 50 =0
the answer is 0
2009-05-05 12:28 am
2cos^2 45°-tan45° /sin^2 50°
= 2(1/sqrt2)^2-1/sin^2 50°
= 1-1/sin^2 50°
= sin^2 50°/sin^2 50°- 1/sin^2 50°
= (sin^2 50°-1)/sin^2 50°
= -(1-sin^2 50°)/sin^2 50°
= -cos^2 50°/sin^2 50°
= -(1/tan^2 50°)(用計算機)
(2cos^2 45°-tan45°) /sin^2 50°
= (2(1/sqrt2)^2-1)/sin^2 50°
= 0/sin^2 50°
= 0
= 2(1/sqrt2)^2-1/sin^2 50°
= 1-1/sin^2 50°
= sin^2 50°/sin^2 50°- 1/sin^2 50°
= (sin^2 50°-1)/sin^2 50°
= -(1-sin^2 50°)/sin^2 50°
= -cos^2 50°/sin^2 50°
= -(1/tan^2 50°)(用計算機)
(2cos^2 45°-tan45°) /sin^2 50°
= (2(1/sqrt2)^2-1)/sin^2 50°
= 0/sin^2 50°
= 0
收錄日期: 2021-04-29 19:19:00
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https://hk.answers.yahoo.com/question/index?qid=20090503000051KK01998