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Trigonometry - Which cofunction identity to use?

  1. Apr 18, 2012 #1
    1. The problem statement, all variables and given/known data
    Write the following in terms of the cofunction identity:
    cos(4pi/3)

    2. Relevant equations
    I know the θ is in Quadrant 3, but my question is, which cofunction identity am I suppose to use?
    Cofunction Identities:
    cosx = sin(pi/2 - x) #1
    cosx = sin(pi/2 + x) #2

    cos(pi + x) = -cosx
    cosx = -cos(pi + x) #3

    cos(pi - x) = -cosx
    cosx = -cos(pi - x) #4

    Is #2 only applicable for angles in quadrant 2 in terms of cofunction identity? But I tried using it and it still gave me the trig ratio. So what is the difference?


    3. The attempt at a solution
    #1 cos(4pi/3) = sin(pi/2 - 4pi/3) = -0.5
    #2 cos(4pi/3) = sin(pi/2 + 4pi/3) = -0.5
    #3 cos(4pi/3) = -cos(pi + 4pi/3) = -0.5
    #4 cos(4pi/3) = -cos(pi - 4pi/3) = -0.5

    See, as shown, they all work.

    I have seen the triangles and know how they are derived, but my main question is, if I am given a question that asks for the equivalent trig ratio in terms of its cofunction identity, is there a specific one I should use when I am doing these types of question?


    Your help is much appreciated.
     
  2. jcsd
  3. Apr 18, 2012 #2

    SammyS

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    Staff Emeritus
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    Gold Member

    They may all be true, but #1 is the cofunction identity.

    It works for all pairs of cofunctions.
    [itex]\displaystyle \cos(x)=\sin\left(\frac{\pi}{2}-x\right)[/itex]

    [itex]\displaystyle \sin(x)=\cos\left(\frac{\pi}{2}-x\right)[/itex]

    [itex]\displaystyle \cot(x)=\tan\left(\frac{\pi}{2}-x\right)[/itex]

    [itex]\displaystyle \tan(x)=\cot\left(\frac{\pi}{2}-x\right)[/itex]

    [itex]\displaystyle \csc(x)=\sec\left(\frac{\pi}{2}-x\right)[/itex]

    [itex]\displaystyle \sec(x)=\csc\left(\frac{\pi}{2}-x\right)[/itex]​
     
  4. Apr 19, 2012 #3
    Thanks for replying Sammy.

    So, the following are identities but they are NOT cofunction identities?
    sin(x+pi/2) = cosx
    cos(x+pi/2) = -sinx
    tan(x+pi/2) = -cotx
    cot(x+pi/2) = -tanx
    csc(x+pi/2) = secx
    sec(x+pi/2) = -cscx

    In this case, my answer for this question would be cos(4pi/3) = sin(pi/2 - 4pi/3) = sin(-5pi/6)? Or do I have to find the related acute of sin(-5pi/6), so it would be sin(pi/6)? But that is not -0.5 though... or is it due to CAST Rule, sin(-5pi/6) is in Quadrant 4, which means sin is negative, and so cos(4pi/3) = sin(-5pi/6) = -sin(pi/6)?

    But then the question arises again, they are all equivalent trig ratios. So which am I suppose to answer with for a question like this? So -sin(pi/6) or sin(-5pi/6) or -cos(pi/3) <- this was the answer my teacher said, after I asked him about this again. But the answer -sin(pi/6) was the answer he told before when he was teaching the lesson.

    Moreover, I would like to ask another question, for the identity: cos(x+pi/2) = -sinx
    Is this only applicable to Quadrant 2 angles? (At least that's what I heard, but I do not understand why, as that would bring you to Quadrant 3, which cos is negative)

    If it is sin(x+pi/2) = cosx, where x is an angle in quadrant 4, then would not cosx be negative and sin be positive? (being in quadrant 1)

    Thank You for your patience and time.
     
  5. Apr 20, 2012 #4

    NascentOxygen

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    Hi reventon_703! To be an identity, equality must hold true for all values of x, with no restrictions or exceptions.

    If you get confused, you can double check a relationship with a calculator. To test cos(x+/2)=sin(x) consider x having a few suitable test values, say, /6 and (+/6). Compare cos(/6+/2) and cos(+/6+/2) and with sin(/6).
     
  6. Apr 21, 2012 #5

    First of all, before I go on, I would like to thank everyone who had view this post or participate in helping with my problem. I had been trying to figure this out for 2 weeks and I had asked my teacher for at least 3 times, yet still in confusion. I have a test on Monday, I understand that I should have post this earlier but I really tried everything out. Examples, googling, most cannot give me an explanation that is concerning to the angles in quadrant 2 and beyond.



    Hi NascentOxygen,

    As you suggested,
    if cos(x + /2) = sin(x), and
    cos(x+/2) = cos(/6+/2) (1),
    cos(x+/2) = cos(+/6 + /2) (2),
    sinx = sin(/6) (3) then:

    1) cos(/6 + /2) NOT= sin(2/3) FALSE
    2) cos(+/6 + /2) = cos(7/6 + /2) = sin(5/3) = -(√3)/2 TRUE
    3) sin(/6) NOT= cos(/6 + /2) FALSE

    Does this means that the identity cos(x + /2) = sinx does indeed applies to angles only in Quadrant 2? But why does my textbook say, cos(x + /2) = -sin(x)??



    Also, can you (or anyone viewing this post) please clarify if the following statements are true?
    #1
    sinθ = cos(θ - /2) applies only to angles in Quadrant 1
    sinθ = cos(θ + /2) applies only to angles in Quadrant 2
    sinθ = cos(θ + ) applies only to angles in Quadrant 3
    sinθ = cos(θ + 3/2) applies only to angles in Quadrant 4

    And that each of the above would work by swapping the sin and cos or switching sin to csc, cos to sec, and sin to tan, cos to cot.

    #2
    For the above statements, which would be negative then? I do not quite understand how cast rule would apply with cofunction, as in the example we worked before sinx = cos(x + /2), if angle (x + /2) is indeed in quadrant 2, would not sin be in quadrant 3 and thus be negative then?

    #3
    What my teacher told me is that you pick the cofunction identity that is "easier" to work in... no idea what he meant by "easier". I questioned why for a Quadrant 4 angle (θ-3/2) would not also work, but he told me that "when you are working with cofunction identity, you are looking at the related acute angle" can you please explain this statement, preferably with an example in syllabus? Would not (θ - 3/2) have worked as long as you find the related acute angle afterwards?

    #4
    If given a question as such cos(4/3) and I were to "rewrite in terms of its cofunction identity", what are the steps involved to solving this? Would it be as such:
    1) Determine the quadrant this angle belongs in - Quadrant 3
    2) Pick the cofunction identity to use - Since it is in Q3, sinθ = cos(θ + ) OR cosθ = sin(θ + )
    3) Substitute - This part I do not understand. Do I split 4/3 to (θ + ) by subtracting ? Or do I merely substitute 4/3 to θ, in which case I would had used a different identity, which is cosθ = sin(θ + )

    Please, any help is appreciate, do reply if you have any thoughts. Your generousity is greatly appreciated as always.
     
  7. Apr 22, 2012 #6

    NascentOxygen

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    Staff: Mentor

    I showed that there is at least one angle where that equation does not hold, so that means it cannot be called an identity.
    Because that is an identity, it holds true for all values of x. Try and prove that it doesn't! :smile:
    You don't need anyone else to investigate these, you can do it yourself. You have a calculator.

    Wiki has a table of identities, should you wish to memorize a few more. http://en.wikipedia.org/wiki/List_of_trigonometric_identities

    BTW, it appears that you were able to view my inclusion of a bold, italic character for Pi () but wherever it should appear in your post I see a pair of ?? on a black diamond background (��). So I'm not sure whether my use is okay or not. I'll keep using it until I hear that some readers cannot view on their browser.
     
  8. Apr 22, 2012 #7
    I had tried this over, and it is true that they work, but my question is really which one to use and why?
    Remember that you told me cos(θ+/2) = -sinx is an identity? So is cos(θ-/2) = sinx, but when would you use each? My teacher told me that you use the first one when you have an angle in quadrant 2, and the second one when you have an angle in quadrant 1. Why? How is it easier?

    Thanks once again for your patience.
     
  9. Apr 22, 2012 #8

    NascentOxygen

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    Staff: Mentor

    This would be easy to apply to an angle in the 2nd quadrant.
    This is cos(-/2+θ) = sinθ so it would be easy to apply to an angle in the 4th quadrant.
     
  10. Apr 22, 2012 #9
    So how exactly is it easier?
    My teacher told me that,
    cos(/2-θ) = sinθ is easier to apply to an angle in 1st quadrant
    cos(/2+θ) = -sinθ is easier to apply to an angle in 2nd quadrant
    cos(+θ) = sinθ is easier to apply to an angle in 3rd quadrant
    cos(3/2+θ) = sinθ is easier to apply to an angle in 4th quadrant

    With the last one, I don't think it even works??
     
  11. Apr 22, 2012 #10

    NascentOxygen

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    Staff: Mentor

    Last edited: Apr 22, 2012
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