三角恆等式求解
回答 (2)
2014-05-20 7:08 pm
✔ 最佳答案
24(Sin2θ+Cos2θ)*60(Sin2π+Cos2π)=?Sol
24(Sin2θ+Cos2θ)*60(Sin2π+Cos2π)
=24(Sin2θ+Cos2θ)*60(0+1)
=1440(Sin2θ+Cos2θ)
2014-05-18 12:56 am
24(sin²θ + cos²θ) × 60(sin2π + cos2π)
=24(1) × 60(sin360° + cos360°)
=24(1) × 60(0+1)
=1440
OR
24(sin²θ + cos²θ) × 60(sin²π + cos²π)
=24(1) × 60(1)
=1440
2014-05-25 13:28:36 補充：
24(Sin2θ+Cos2θ)x60(Sin2π+Cos2π)
=24(Sin2θ+Cos2θ)x60(0+1)
=1440(Sin2θ+Cos2θ)
=24(1) × 60(sin360° + cos360°)
=24(1) × 60(0+1)
=1440
OR
24(sin²θ + cos²θ) × 60(sin²π + cos²π)
=24(1) × 60(1)
=1440
2014-05-25 13:28:36 補充：
24(Sin2θ+Cos2θ)x60(Sin2π+Cos2π)
=24(Sin2θ+Cos2θ)x60(0+1)
=1440(Sin2θ+Cos2θ)
收錄日期: 2021-04-13 22:22:48
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140517000051KK00097