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Homework Help: When do I use sin or cos?

  1. May 12, 2014 #1
    Not a direct problem but this is a homework related question so I'm posting it here. When getting components with respect to gravity it is often intuitive, this isn't always the case. None of my classes thus far have talked about why when to use sin or cos, they just do. I'm wondering how I can figure this out when the answer isn't obvious.
  2. jcsd
  3. May 12, 2014 #2


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    Homework Helper

    You can use either as long as your use of them is consistent (because sin(θ) = cos(θ+90°) so it just depends on how you define your coordinates)

    It's hard to abstractly explain which to use without a specific question, but if it confuses you then you should re-study some trigonometry. (It's a lack of understanding of trigonometry that makes it confusing, not a lack of understanding of physics.)
  4. May 12, 2014 #3
    They don't seem interchangeable. Here's a specific problem I just did (and got right). Is the way I used trig towards the top correct (I reasoned that the component passes though the adjacent and hypotenuse so it must be cos)?

  5. May 12, 2014 #4


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    Staff: Mentor

    As you were told, it all depends on how you define your coordinates and where does your angle start (or how it is measured). It doesn't mean sin and cos are interchangeable.

    No idea what is given and what you are trying to calculate, so hard to comment on what you did.
  6. May 12, 2014 #5
    wow I just spent 10 minutes writing a reply explaining everything and I accidently closed the tab. Basically your teacher was just decomposing the forces into their x and y components to apply newton's laws. If you have a right angled triangle with angles [itex]\phi[/itex] and [itex]\theta[/itex] then the cos of [itex]\theta[/itex] = the sin of [itex]\phi[/itex]. They are interchangeable.

    watch this https://www.khanacademy.org/math/tr...ometry/basic_trig_ratios/v/basic-trigonometry
  7. May 13, 2014 #6


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    Homework Helper

    which angle is called theta (θ)?

    If it is the angle of the beam with respect to the horizontal the equation for the torque ( τ ) is correct. Torque is force multiplied with the lever of arm, the distance of the line of force from the rotation axis.

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