Express 4 sin 2x – 3cos 2x in the form of
Rcos(2x +α), α>=0°
. Hence
find the smallest positive value of x for which 4 sin 2x – 3cos 2x = 4.
i can't Calculate help me?
2015-10-16 4:03 pm
回答 (2)
2015-10-16 5:55 pm
4 sin 2x - 3 cos 2x = R cos [ 2x - α ]
4 sin 2x - 3 cos 2x = R cos α [cos 2x] - R sin α [sin 2x]
- 3 = R cos α
- 4 = R sin α
tan α = 4/3 ________α in 3rd quadrant
α = 233.1⁰
9 + 16 = R²
R = 5
5 cos [ 2x + 233.1 ]⁰ = 4
cos [ 2x + 233.1 ]⁰ = 4/5
[ 2x + 233.1 ]⁰ = 36.9⁰
2x = - 196 2⁰
x = - 93.1 ⁰
x = 266.9 ⁰
4 sin 2x - 3 cos 2x = R cos α [cos 2x] - R sin α [sin 2x]
- 3 = R cos α
- 4 = R sin α
tan α = 4/3 ________α in 3rd quadrant
α = 233.1⁰
9 + 16 = R²
R = 5
5 cos [ 2x + 233.1 ]⁰ = 4
cos [ 2x + 233.1 ]⁰ = 4/5
[ 2x + 233.1 ]⁰ = 36.9⁰
2x = - 196 2⁰
x = - 93.1 ⁰
x = 266.9 ⁰
2015-10-16 4:22 pm
well cos (x + pi/2) = sin(x)
4sin2x = 4cos(2x + n(pi/2))
4cos(2x +n( pi/2)) + 3 cos(2x)
4sin2x = 4cos(2x + n(pi/2))
4cos(2x +n( pi/2)) + 3 cos(2x)
收錄日期: 2021-04-21 14:40:56
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151016080315AAFmXIc