1 / 4

Example

Example. Express -8sinx° + 15cosx° in the form ksin(x +  )°. *********. a = -8 & b = 15. kcos  °<0 Q2 or Q3. kcos  ° = -8 and ksin  ° = 15. ksin  °>0 Q1 or Q2. (kcos  °) 2 + (ksin  °) 2 = (-8) 2 + 15 2. k 2 = 64 + 225. . . . k 2 = 289. . k = 17.

gyda
Download Presentation

Example

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. Example Express -8sinx° + 15cosx° in the form ksin(x + )° ********* a = -8 & b = 15 kcos°<0 Q2 or Q3 kcos° = -8 and ksin° = 15 ksin°>0 Q1 or Q2 (kcos°)2 + (ksin°)2 = (-8)2 + 152 k2 = 64 + 225    k2 = 289  k = 17 tan° = b/a = 15/-8  in Q2 tan-1(15/8) = 61.9° Q2:  = 180 – 61.9 = 118.1° So -8sinx° + 15cosx° = 17sin(x + 118.1)°

  2. Example (radians) Express 3/2sin - 1/2cos in the form ksin( + ) ********* a = 3/2 & b = -1/2 kcos>0 Q1 or Q4 kcos = 3/2 and ksin = -1/2 ksin<0 Q3 or Q4 (kcos)2 + (ksin)2 = (3/2 )2 + (-1/2)2 k2 = 3/4 + 1/4  k2 = 1    k = 1  in Q4 tan = b/a = (-1/2)(3/2) = -1/2X ( 2/3) = -1/3 tan-1 (1/3)= 30° = /6 Q4:  = 2 - /6 = 11/6 So 3/2sin - 1/2cos = sin( + 11/6)

  3. VARIATIONS The usual formats are acosx° + bsinx° = kcos(x - )° asinx° + bcosx° = ksin(x + )° The variations kcos(x + )° and ksin(x - )° are easily obtained by considering reverse rotations from the X-axis. NB: -200° = +160° while +120° = -240° So sin(x + 225)° = sin(x – 135)° and cos( - 7/5) = cos( + 3/5)

  4. Example Express 12sinx° - 5cosx° in the form ksin(x - )° ********* a = 12 & b = -5 kcos° = 12 and ksin° = -5 kcos°>0 Q1 or Q4 (kcos°)2 + (ksin°)2 = 122 + (-5)2 ksin°<0 Q3 or Q4 k2 = 144 + 25 k2 = 169  k = 13    tan° = b/a = -5/12  in Q4 tan-1(5/12) = 22.6° Q4:  = 360 – 22.6 = 337.4° So 12sinx° - 5cosx° = 13sin(x + 337.4)° = 13sin(x – 22.6)°

More Related