1 / 4

SOLVING WAVE EQUATIONS

SOLVING WAVE EQUATIONS. In a previous example we saw that 7cosx° + 24sinx° = 25cos(x – 73.7)°. So the equation 7cosx° + 24sinx° = 15. ( 25). Could be changed to 25cos(x – 73.7)° = 15. cos(x – 73.7)° = 0.6. Q1 or Q4. cos -1 0.6 = 53.1°. Q1: angle = 53.1° so.

lew
Download Presentation

SOLVING WAVE EQUATIONS

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. SOLVING WAVE EQUATIONS In a previous example we saw that 7cosx° + 24sinx° = 25cos(x – 73.7)° So the equation 7cosx° + 24sinx° = 15 (25) Could be changed to 25cos(x – 73.7)° = 15 cos(x – 73.7)° = 0.6 Q1 or Q4 cos-10.6 = 53.1° Q1: angle = 53.1° so x – 73.7° = 53.1° ie x° = 126.8° Q4: angle = 360° – 53.1° so x – 73.7° = 306.9° ie x = 380.6° (-360) ie x = 20.6° Check: 7cos20.6° + 24sin20.6° = 14.9966…  15

  2. Example Solve 5cosx° – 12sinx° + 6.5 = 0 where 0<x<360 ******** Let 5cosx° – 12sinx° = kcos(x - )° kcos°>0 Q1 or Q4 a = 5 & b = -12 ksin°<0 Q3 or Q4 kcos° = 5 and ksin° = -12  (kcos°)2 + (ksin°)2 = 52 + (-12)2    k2 = 25 + 144 k2 = 169  in Q4 k = 13 tan° = b/a = -12/5 tan-1(12/5) = 67.4° Q4:  = 360° – 67.4° = 292.6° So 5cosx° – 12sinx° = 13cos(x – 292.6)°

  3. So 5cosx° – 12sinx° + 6.5 = 0 becomes 13cos(x – 292.6)° + 6.5 = 0 13cos(x – 292.6)° = -6.5 cos(x – 292.6)° = -0.5 Q2 or Q3 cos-10.5 = 60° Q2: angle = 180° – 60° so x – 292.6° = 120° or x = 412.6° (-360) ie x = 52.6° Q3: angle = 180° + 60° so x – 292.6° = 240° or x = 532.6° (-360) ie x = 172.6° Check 5cos172.6° - 12sin172.6° + 6.5 = -0.0039…  0

  4. Example Solve 3sin + cos + 2 = 0 where 0<  <2 ******** Let 3sin + cos = ksin( + ) kcos>0 Q1 or Q4 a = 3 & b = 1 ksin>0 Q1 or Q2 kcos° = 3 and ksin° = 1    (kcos)2 + (ksin)2 = (3)2 + 12  k2 = 3 + 1 k2 = 4  in Q1 k = 2 tan° = b/a = 1/3 3sin + cos = 2sin( + /6) Q1:  = tan-1(1/3) = 30° = /6 So 3sin + cos + 2 = 0 becomes 2sin( + /6) + 2 = 0 consider graph sin( + /6) = -1 ie  = 4/3 angle = 270° = 3/2 so  + /6 = 3/2 so  = 3/2 - /6

More Related