更新1:
find the general solution of cos x - √3sin x = 1 (in degrees)
有a.maths 唔識做
回答 (3)
2008-06-08 4:17 am
✔ 最佳答案
sinθ + cos θ = √2 (sinθ + cos θ )/√2=1
cosθsinθ+sinθcosθ=1, where cosθ=1/√2
2sinθcosθ=1
sin2θ=1
2θ=180°n+(-1)^n(90°)
θ=90°n+(-1)^n(45°)
cos x - √3sin x = 1
1/2cosx-√3/2sinx=1/2
sin30°cosx-cos30°sinx=1/2
sin(30°-x)=sin30°
30°-x=180°n+(-1)^n(30°)
x=30°-180°n+(-1)^n(30°)
2008-06-09 2:06 am
第一條最簡單做法係:
sinx+sin(90-x)=√2
sinθ+sinθ=√2
2sinθ=√2
sinθ=√2/2
sinθ=sin90
θ=180+(-1)^n+90
sinx+sin(90-x)=√2
sinθ+sinθ=√2
2sinθ=√2
sinθ=√2/2
sinθ=sin90
θ=180+(-1)^n+90
2008-06-08 4:16 am
sinθ + cos θ = √2
√2 [sinθ (1/√2 )+cosθ( (1/√2 )]=√2
√2 sin(θ+45˚)=√2
sin(θ+45˚)=1
θ+45˚=180n˚+(-1)n(90˚),where n is an integer.
θ=180n˚+(-1)n(90˚)+45˚
2008-06-07 20:21:58 補充：
cos x - √3sin x = 1
2[cos x(1/2) - sin x(√3/2)] = 1
2cos(x+60˚)=1
cos(x+60˚)=1/2
x+60˚=360n˚±60˚,where n is an integer.
x=360n˚+120˚ or x=360n˚
√2 [sinθ (1/√2 )+cosθ( (1/√2 )]=√2
√2 sin(θ+45˚)=√2
sin(θ+45˚)=1
θ+45˚=180n˚+(-1)n(90˚),where n is an integer.
θ=180n˚+(-1)n(90˚)+45˚
2008-06-07 20:21:58 補充：
cos x - √3sin x = 1
2[cos x(1/2) - sin x(√3/2)] = 1
2cos(x+60˚)=1
cos(x+60˚)=1/2
x+60˚=360n˚±60˚,where n is an integer.
x=360n˚+120˚ or x=360n˚
收錄日期: 2021-04-23 20:33:13
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https://hk.answers.yahoo.com/question/index?qid=20080607000051KK02686