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SOLVING WAVE EQUATIONS

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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.

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SOLVING WAVE EQUATIONS

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  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

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