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Understanding when to use sine and cosine to find x and y components

  1. Sep 15, 2007 #1
    I am having trouble understanding when to use sine and cosine to find x and y components. I know that its not always going to be the same (ex. you won't always use cosine to find x component.) Any input would be appreciated!
     
  2. jcsd
  3. Sep 15, 2007 #2

    cristo

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    If you have a vector, then the horizontal and vertical components of that vector are the sides of the right angled triangle that has that vector as the hypotenuse. Then, to find the components, you use trigonometry. You will know the angle between the x axis and the vector, and you know the magnitude of the vector, so you will be able to find the horizontal and vertical components using the familiar relations sin(theta)=opposite/hypotenuse and cos(theta)=adjacent/hypotenuse.
     
  4. Sep 15, 2007 #3

    mjsd

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    mmm... if you are confused, first draw a picture. It always works for me. then identify the opposite and adjacent side of the triangle formed by your vector and the x, y-axis given the angle that you know. that way you will know whether to use sin or cos.
     
  5. Sep 15, 2007 #4
    I understand what you guys are saying but look at this...

    The Problem:
    Three forces in the x-y plane act on a 3.70 kg mass: 14.50 N directed at 478.0°, 11.00 N directed at 117.0°, and 10.70 N directed at 222.0°. All angles are measured from the positive x-axis, with positive angles in the counter-clockwise direction. Calculate the magnitude of the acceleration.

    What I did:
    Fx1= 14.5 sin 478
    Fx2= 11.0 sin 117
    Fx3= 10.7 cos 222
    sum= 14.65216223

    Fy1= 14.5 cos 478
    Fy2= 11.0 cos 117
    Fy3= 10.7 sin 222
    sum= -18.96093065

    But what you are supposed to do is use cos for x comp. and sin for y comp. but they don't match up that way when using the trig functions.
     
  6. Sep 15, 2007 #5

    cristo

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    The easiest way (in my opinion) to tackle that sort of question is to first draw a diagram. Then, get rid of all the horrible angles, and use the corresponding angle between 0 and 90. i.e. the first vector is 67 degrees clockwise from the negative x axis. Then you will find it easier to resolve this vector into x and y components.


    [Isn't your quote from Einstein, mjsd?]
     
  7. Sep 15, 2007 #6

    mjsd

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    yeah! draw a diagram :smile:
    but since all angles are measured from +ve x-axis anyway, things are so easy that you just plug in the angle: cos for x-comp, sin for y-comp. what do u mean by they don't match up??
     
  8. Sep 15, 2007 #7
    Yea I always draw a diagram but if you draw out those angles and vectors some of the require sine to find x-component and 1 requires cosine because its in another quadrant.
     
  9. Sep 15, 2007 #8

    mjsd

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    no, you don't. if you angle is defined from the +ve x-axis always, then cos will always give you the projection onto the x-axis (ie. x-comp) and sin will always give the y-comp.
    eg. vector of length 1 angle is 120, that in the 2nd quadrant BUT the x-comp is still given by cos 120 = -0.5 you get a negative number.... reflecting the fact that you are in the 2nd quadrant. the confusion comes from the fact that you think the angle (120) is too large, and you have to take away 180 from it to get the acute angle.. but the point is that cosine function will automatically take care of that . If you evaluate instead
    cos (180-120) = 0.5 (which is wrong.. and you need to put in a -ve sign yourself to fix it because the angle (180-120) is measured from the -ve x-axis. and if you use sin (180-120) or sin (120) that's absolutely wrong)

    because in this case all angles are given relative to the +ve x-axis (going clockwise), you don't have to worry about anything at all. cos -> x-comp, sin -> y-comp.
     
    Last edited: Sep 15, 2007
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