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Equations & Inequalities

  1. Jul 27, 2008 #1
    Hello Guys, I'm new to these boards, just wanted to say hi before I start my post.

    I had some questions about solving equations and inequalities which I wasn't sure how to do...




    Find all solutions for [tex]\theta \in [0, 2\pi][/tex]

    (I) [tex]sin(\theta - \frac{\pi}{4}) = \frac{\sqrt{3}}{2}[/tex]
    (II) [tex]tan(3 \theta + \frac{\pi}{6}) = \frac{-1}{\sqrt{3}}[/tex]

    Attemp at a solution;
    (I)
    [tex]sin(\theta - \frac{\pi}{4}) = \frac{\sqrt{3}}{2}[/tex]
    [tex]\theta - \frac{/pi}{4} = sin^{-1} (\frac{\sqrt{3}}{2})[/tex]
    [tex]\theta - \frac{/pi}{4} = 60[/tex]
    [tex]\theta = 60 + (\frac{/pi}{4})[/tex]
    [tex]\theta = 60.78[/tex]

    But what does the question mean by "find all solutions"? Does this mean there any more solutions to this? :uhh:


    (II)
    [tex]tan(3 \theta + \frac{\pi}{6}) = \frac{-1}{\sqrt{3}}[/tex]
    [tex]3 \theta + \frac{\pi}{6} = tan^{-1} \frac{-1}{\sqrt{3}}[/tex]
    [tex]3 \theta = 29.3 - \frac{\pi}{6}[/tex]
    [tex]\theta = -27.08 /3 = -9.02[/tex]

    I'm a new student so I'm not very familiar with the question.. I appreciate any one who can help me...

    P.S.
    Do I need to have my calculator set degrees or radians for such questions?

     
  2. jcsd
  3. Jul 27, 2008 #2

    HallsofIvy

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    No! [itex]sin^{-1}(-\sqrt{3}/2[/itex] is not "60" it is [itex]\pi/6[/itex]
    Of course 60 degrees is the same as [itex]\pi/6[/itex] radians but here the answer must be in radians. That is because in the function f(x)= sin(x), x is not, strictly speaking, an angle at all. There are a number of different ways to define sin(x), etc. so that they apply to any value. The standard trigonometry definition, using right angles, restricts x to lie between 0 and 90 degrees (or 0 and [itex]\pi/2[/itex] radians) or, in more general triangles, between 0 and 180 degerees (0 and [itex]\pi[/itex]) radians. We need other definitions which extend x to all real numbers. I don't know which your textbook uses but saying that cos(t) and sin(t) are the x, y, components of a point on the unit circle at distance t around the circumference of the circle, in the count-clockwise direction, from (1, 0) is perhaps most common. Notice that t, here, is not an angle at all. It is a length, just as x and y are lengths along the axes. Of course, radian measure is defined so that [itex]\theta[/itex] radians exactly corresponds to a measure of [itex]\theta[/itex] on the circumference of the unit circle. That is also "signaled" to you by the fact that, in the problem itself, [itex]\pi/4[/itex] is added to x.


    As I said before, your equation should be [itex]\theta- \pi/4= \pi/6[/itex] so [itex]\theta= \pi/4+ \pi/6= 3\pi/12+ 2\pi/12= 5\pi/12[/itex] is you "base" angle.

    Yes, of course. You know, I hope, that the "trig" functions are periodic with period [itex]2\pi[/itex] (because going around the circumference of the unit circle an additional distance [itex]2\pi[/itex] takes you exactly once around the circumference, putting you right back at the same point with the same x,y values. So if [itex]\pi/6[/itex] is a solution to [itex]sin(x)= \sqrt{2}/2[/itex], so is [itex]\pi/6+ 2n\pi[/itex] for any integer n. But more than that: since sin(x) is the y coordinate of a point on the unit circle, we can see that a y= constant line, in particular, [itex]y= \sqrt{2}/2[/itex] crossses the unit circle twice. For y positive (and less than or equal to 1), the horizontal line crosses the unit circle at [itex]\theta[/itex] and [itex]\pi - \theta[/itex]. If [itex]\theta= \pi/6[/itex] is a solution, then so is [itex]\pi- \theta= \pi- \pi/6= 5\pi/6[/itex]. And so, then, is [itex]5\pi/6+ 2n\pi[/itex] for any integer n.

    Same criticisms: you should be working in radians, not degrees and, I suspect, for these problems you should be using exact values, not "calculator" values. Also, tan(x) is periodic with period [itex]\pi[/itex].

    Finally, you ask! Unless the problems are specifically "triangle" problems, with angles given in degrees, you should use radians. However for the special values given here you should not use a calculator at all- these have exact values.
     
  4. Jul 27, 2008 #3
    Thanks. I appreciate your help! :smile: I made some notes in my book, but just to clear up somethings I didn't understand;

    In [tex]tan(3 \theta + \frac{\pi}{6}) = \frac{-1}{\sqrt{3}}[/tex] how can I find [tex]tan^{-1} (\frac{-1}{\sqrt{3}})[/tex] using exact values? (When I use my calculator I get -0.52).

    But you're right, it has to be in radians and in exact values!


    I just write the first question again, please correct me if it's wrong;

    [tex]sin(\theta - \frac{\pi}{4}) = \frac{\sqrt{3}}{2}[/tex]
    [tex]\theta - \frac{\pi}{4} = \frac{\pi}{6}[/tex]

    [tex]\theta= \frac{5\pi}{12}[tex]

    So [tex]\frac{5\pi}{12}[/tex] is a solutions and we have to add multiples of [tex]2n\pi[/tex] to it, where n = ...-1, 0, 1, 2... as long as all the solutions lie within [tex]\theta \in [0, 2\pi][/tex]. Is that correct?


    Have a nice day.
     
    Last edited: Jul 27, 2008
  5. Jul 27, 2008 #4

    tiny-tim

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    Welcome to PF!

    Hi powerless! Welcome to PF! :smile:

    You need to learn your trigonometric identities. :wink:

    In this case, use sec2x = tan2x + 1, so tan-1(-1/√3) = sec-1(…) = cos-1(…) ? :smile:
     
  6. Jul 27, 2008 #5
    Hello Comrade,

    I didn't understand sorry :shy:

    e.g. how did you figure [tex]sin^{-1}(\sqrt{3}/2)[/tex], which is equal to 60°, is equal to [tex]\pi/6[/tex]? (in radians & exact value)
    Because [tex]\pi = 180[/tex] right? so [tex]\frac{\pi}{6} = 30[/tex] not 60.
    If I get this one I can do the other one on my own. I’m so overwhelmed with stress right now.
    I have to learn all these since I’ve got a few days left before my exam and I’m starting to choke. :(
     
    Last edited: Jul 27, 2008
  7. Jul 27, 2008 #6

    tiny-tim

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    Hi powerless! :smile:

    I'm confused … where does [tex]sin^{-1}(\sqrt{3}/2)[/tex] come from? :confused:

    Isn't it [tex]cos^{-1}(\sqrt{3}/2)[/tex] ? :smile:

    (btw, see trigonometric identities)
     
  8. Jul 28, 2008 #7
    No, tiny tim!

    I was trying to solve two different equations: [tex]sin(\theta - \frac{\pi}{4}) = \frac{\sqrt{3}}{2}[/tex] and [tex]tan(3 \theta + \frac{\pi}{6}) = \frac{-1}{\sqrt{3}}[/tex]

    For example when solving the second one, I don't know what [tex]tan^{-1} (\frac{-1}{\sqrt{3}})[/tex] equals to.
    I can use my calculator and by using that I get -0.52. But we need exact values (in radians) not calculator values. :frown:

    Like in the first equation: [tex]sin^{-1}(\sqrt{3}/2)[/tex] = [tex]\pi/6[/tex]
     
  9. Jul 28, 2008 #8

    Defennder

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    You need to look up a trigo table for that. Google that.
     
  10. Jul 28, 2008 #9

    HallsofIvy

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    Draw an equilateral triangle with sides of length 2. All angles are 60 degrees or [itex]pi/34[/itex] radians. Drop a perpendicular from the vertex to the base. That divides the equilateral triangle into two right triangles have 30 degree ([itex]pi/6[/itex]) and 60 degree ([itex]\pi/3[/itex]) angles, hypotenuse of length 2 and one leg of length 1. Use the Pythagorean theorem to determine that the length of the other leg is [itex]\sqrt{2^2- 1}= \sqrt{3}[/itex]. Now it is easy to see that [itex]sin(\pi/3)= \sqrt{3}/2[/itex] and [itex]cos(\pi/3)= 1/2[/itex] as well as [itex]sin(\pi/6)= 1/2[/itex] and [itex]cos(\pi/6)= \sqrt{3}/2[/itex].

    Other angles: Draw an isosceles right triangle with legs of length 1. Two of the angles are 45 degrees ([itex]\pi/4[/itex] and the hypotenuse has length, again by the Pythagorean theorem, [itex]\sqrt{2}[/itex]. From that [itex]sin(\pi/4)= cos(\pi/4)= 1/\sqrt{2}= \sqrt{2}/2[/itex].
     
  11. Jul 28, 2008 #10
    OK, What about the first one? is it right?

    [tex]sin(\theta - \frac{\pi}{4}) = \frac{\sqrt{3}}{2}[/tex]

    [tex]\theta - \frac{\pi}{4} = \frac{\pi}{6}[/tex]

    [tex]\theta= \frac{5\pi}{12}[/tex] is a solutions and we have to add multiples of [tex]2n\pi[/tex] to it, where n = ...-1, 0, 1, 2... as long as all the solutions lie within [tex]\theta \in [0, 2\pi][/tex]
     
  12. Jul 28, 2008 #11

    HallsofIvy

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    Well, adding multiples of [itex]2\pi[/itex] will take it out of [itex][0, 2\pi][/itex]. There are only two solutions in [itex][0, 2\pi][/itex]
     
  13. Jul 29, 2008 #12

    [tex]sin(\theta - \frac{\pi}{4}) = \frac{\sqrt{3}}{2}[/tex]

    [tex]\theta - \frac{\pi}{4} = \frac{\pi}{6}[/tex]

    [tex]\theta= \frac{5\pi}{12}[/tex] *a solution*

    so what should I do? can I add multiples of [tex]\pi[/tex] instead?

    [tex]\frac{5\pi}{12} + \pi[/tex] = [tex]\frac{\pi}{2}[/tex]

    Is [tex]\frac{\pi}{2}[/tex] the second solution?
     
  14. Jul 29, 2008 #13
    I would suggest learning more about the unit circle. The second solution follows from symmetry on the unit circle. More specifically, the identity sin(180-x) = sin(x) can be justified with a simple geometric argument. But learn the unit circle first.

    This helped a lot when I first started learning trig:

    http://www.stanford.edu/~lambers/math1b/lecture7.pdf
     
  15. Jul 29, 2008 #14
    Hi!

    So, [tex]\pi - \frac{5\pi}{12}[/tex] = [tex]\frac{-4\pi}{12}[/tex] = [tex]\frac{-\pi}{3}[/tex]
    Is the 2nd solution, right?

    I'm really trying to learn the unit circle but I haven't quite grasped the idea just yet.
     
  16. Jul 29, 2008 #15
    Ok, there is a glaring error, but relax. Don't worry about the exam but focus on basic principles. First off, read HallOfIvy's post on trigonometric values carefully. You need to know those by heart before you can begin to solve this problem, let alone understand the unit circle. Also recall the radian/degree conversion factor pi/180 = 1.

    Once you have those values down, consider the equation again. The right hand side sqrt(3)/2 should look very familiar now. It is sin(pi/3) = cos(pi/6) [NOT sin(pi/6) = 1/2, which is why we have to start over]. Thus, our equation becomes sin(x - pi/4) = sin(pi/3). Now you can find one solution pretty easily, but of course it's not the only one.

    So how do we apply sin(pi - x) = sin(x)? First of all, find the first solution for x above. Plug x back in to make sure it checks out (you should ALWAYS do this), i.e. the left hand side is sin(pi/3) as well. Now can you apply sin(pi -x) = sin(x)? Can you then find the other value of x on [0, 2pi]?
     
  17. Jul 29, 2008 #16
    [tex]sin(\theta - \frac{\pi}{4}) = \frac{\sqrt{3}}{2}[/tex]


    Why does it become sin(x - pi/4) = sin(pi/3)??

    When solving for [tex]\theta[/tex], don't we have to take [tex]Sin^{-1}[/tex] of the other side?
    [tex]sin^{-1}(-\sqrt{3}/2)[/tex] [itex]= \pi/3[/tex] I looked at the table and that's what it says.

    [tex]\theta - \frac{\pi}{4} = \frac{\pi}{3}[/tex]
    [tex]\theta = \frac{\pi}{3} + \frac{\pi}{4}[/tex]
    [tex]\theta = \frac{7 \pi}{12}[/tex]

    This is one of the things I must know for my exam but everyone is saying something different. At some points during the day I feel confident with my knowledge in mind, but at other times I start to shake. What if I fail? OMG
     
    Last edited: Jul 29, 2008
  18. Jul 31, 2008 #17
    please help!

    It's very difficult for me because i don't have any worked examples here to assist me.

    I will attempt this one:
    [tex]tan(3 \theta + \frac{\pi}{6}) = \frac{-1}{\sqrt{3}}[/tex]

    [tex](3 \theta + \frac{\pi}{6}) = \frac{5\pi}{6}[/tex]

    [tex]3 \theta = \frac{5\pi}{6}-\frac{\pi}{6}[/tex]

    [tex]\theta = \frac{4\pi}{6} . \frac{1}{3}[/tex]

    [tex]\theta = \frac{2\pi}{9}[/tex]

    Now to find all solutions

    [tex]\pi - \frac{2\pi}{9} = \frac{\pi}{9}[/tex]

    & [tex]\pi + \frac{2\pi}{9} = \frac{3\pi}{9} = \frac{\pi}{3}[/tex]

    Is any of that correct?
     
  19. Jul 31, 2008 #18

    tiny-tim

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    Hi powerless! :smile:

    hmm …
    [tex]tan(3 \theta + \frac{\pi}{6}) = \frac{-1}{\sqrt{3}}[/tex]

    [tex](3 \theta + \frac{\pi}{6}) = \frac{5\pi}{6}[/tex] + nπ

    [tex]3 \theta = \frac{5\pi}{6}-\frac{\pi}{6}[/tex] + nπ

    [tex]\theta = \frac{4\pi}{6} . \frac{1}{3}[/tex] + nπ/3

    So [tex]\theta = \frac{2\pi}{9}[/tex] or … or … ? :smile:
     
  20. Jul 31, 2008 #19
    Hi Tiny Tim!

    OK I do it again;

    [tex]\theta = \frac{4\pi}{6} . \frac{1}{3} + \frac{n\pi}{3}[/tex]


    [tex]\theta = \frac{2\pi}{9}+\frac{n \pi}{3}[/tex]

    The solutions must be within [0, 2π]

    If n = 0 [tex]\theta = \frac{2\pi}{9}[/tex]

    If n = 1 [tex]\theta = \frac{2\pi}{9}+\frac{\pi}{3} = \frac{5 \pi}{9}[/tex]

    If n = 2 [tex]\theta = \frac{2\pi}{9} + \frac{2 \pi}{3} = \frac{8 \pi}{9}[/tex]

    If n = 3 [tex]\theta = \frac{2\pi}{9} + \frac{3 \pi}{3} = \frac{11 \pi}{9}[/tex]

    Do I need to continue like this or not? Is it right so far?
     
  21. Jul 31, 2008 #20

    tiny-tim

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    Hi powerless! :smile:

    Yes, that looks fine! :smile:

    But you don't have to keep mentioning the n's (they're your n's, they're not part of the question).

    So just say:

    [tex]\theta = \frac{2\pi}{9}+\frac{n \pi}{3}[/tex]

    The solutions must be within [0, 2π]​

    and then go straight to listing them. :smile:
     
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