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Homework Help: Need help with sec, sin, etc.

  1. Feb 14, 2008 #1
    1. The problem statement, all variables and given/known data

    4sec180° - 2sin2270°

    3. The attempt at a solution

    sec(180 is -1.

    So we have 4(-1) which is -4

    sin2(270 is -1
    So we have 2(-1)

    This now reads -4 - -2

    Answer = -2

    Is this correct or do I suck at this? =(
  2. jcsd
  3. Feb 14, 2008 #2


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    Dearly Missed

    What sign must the square of a real number always have?
  4. Feb 14, 2008 #3
  5. Feb 14, 2008 #4


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    Yes. sin(270)= -1 so sin2(270)= (-1)2= (-1)(-1)= 1.
  6. Feb 14, 2008 #5
    Ok let me try this again:

    The attempt at a solution

    sec(180 is -1.

    So we have 4(-1) which is -4

    sin2(270 is (-1)2 which is 1

    So we have 2(1)

    This now reads -4 - 2

    Answer = -6

    Is this right?
    Last edited: Feb 14, 2008
  7. Feb 14, 2008 #6
    Ok here is another question, a bit off topic. I know that sec(180 = -1. But this is the question I have.

    -When looking at 180° in standard position, we realize that y is 0 and x is any given number along the x axis.
    -sec is understood to be r/x.

    - So let's say I choose (5,0)
    -I need to find r, so we do square root of (5)2 + (0)2
    -Which leads to the square root of 25
    -Squared is 5

    Therefore 5/5 = 1 and sec=1

    My question is why in the calculator it says sec(180 is -1? Is it because 180° lies in both the negative x quadrants (II and III) and the postive x quadrants (I and IV?) Since it lies in both a negative and positive x axis, it must be -1 instead of 1? This confuses me a little bit, I know I probably sound like a moron for even asking such a question.
    Last edited: Feb 14, 2008
  8. Feb 15, 2008 #7

    Gib Z

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    Remember [tex]\sec x = \frac{1}{\cos x}[/tex] by definition. Do you remember the definitions of the trig functions on the unit circle? If you do, Cos 180° should be quite easy seen to be 1 =]
  9. Feb 15, 2008 #8
    Plot your (5,0) point on a graph. Draw an arrow from the origin to the point. Which direction (degrees) does the arrow point?
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