一條咁多 MATHS
2010-04-25 6:34 am
回答 (3)
2010-04-25 6:46 am
✔ 最佳答案
7sinθ+cosθ = 2(4sinθ+cosθ)sinθ+cosθ=0
(√2)cosθcos45°+(√2)sinθsin45°=0
(√2)cos(θ−45°)=0
cos(θ−45°)=0
θ−45°=360°n±90°
θ = 360°n+135° or 360°n−45° ,where n is any integer.
2010-04-28 2:17 am
(7sin@ + cos@) / (4sin@ + cos@) = 2
7sin@ + cos@ = 2(4sin@ + cos@)
7sin@ + cos@ = 8sin@ + 2cos@
sin@ = -cos@
tan@ = -1
@ = 135* or 315*
for 0 <= @ <= 360*
7sin@ + cos@ = 2(4sin@ + cos@)
7sin@ + cos@ = 8sin@ + 2cos@
sin@ = -cos@
tan@ = -1
@ = 135* or 315*
for 0 <= @ <= 360*
2010-04-27 4:53 am
For the answer, I think 002 is more comprehensive. He mentions all solutions can be obtained called general solutions
收錄日期: 2021-04-19 22:02:49
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100424000051KK01754