中四數學題

2015-07-27 9:01 pm
1. sin(270 ° - θ ) / sin θ + cos(90 °+ θ ) / cos θ

2. 4sin(90 °- θ ) * sin(θ-360°) + cos (270°-θ )



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回答 (2)

2015-07-27 10:06 pm
✔ 最佳答案

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1.
sin(270° - θ) / sin(θ) + cos(90° + θ) / cos(θ)
= sin(180° + 90° - θ) / sin(θ) + cos[180° - (90° - θ)] / cos(θ)
= -sin(90° - θ) / sin(θ) - cos(90° - θ) / cos(θ)
= -cos(θ) / sin(θ) - sin(θ) / cos(θ)
= -[ cos(θ) / sin(θ) + sin(θ) / cos(θ) ]
= -{ [cos²(θ) + sin²(θ)] / [sin(θ) cos(θ)] }
= -1/[sin(θ) cos(θ)]

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2.
4sin(90° - θ) × sin(θ - 360°) + cos(270° - θ)
= 4cos(θ) × [-sin(360° - θ)] + cos(180° + 90° - θ)
= -4cos(θ) × sin(360° - θ) - cos(90° - θ)
= -4cos(θ) × [-sin(θ)] - sin(θ)
= 4sin(θ)cos(θ) - sin(θ)
= sin(θ)[4cos(θ) - 1]
2015-07-29 3:59 am
1.
sin(270 ° - θ ) / sin θ + cos(90 °+ θ ) / cos θ
=-cos θ /sin θ -sin θ /cos θ
=-[(1/tan θ )+tan θ ]
=-[1+(tan θ )(tan θ )]/tan θ

2.
4sin(90 °- θ ) * sin(θ-360°) + cos (270°-θ )
=4cos θ sin θ -sin θ
=sin θ (4cos θ -1)


收錄日期: 2021-04-18 00:11:02
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150727000051KK00032

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