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Homework Help: General solution of trig functions

  1. Sep 20, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the general solution of:
    a) sin x = 1/[tex]\sqrt{2}[/tex]
    b) cos x = 0.5

    3. The attempt at a solution
    a) x = asin(1/[tex]\sqrt{2}[/tex])
    x = (π/4) + 2nπ

    b) x = acos(0.5)
    x = π/3 + 2nπ

    Basically, my strategy was to solve for the basic angle, and then add multiples of the period (2nπ for sin and cos, nπ for tan).

    However, the answers provided are:

    a) x = 2nπ + asin(1/[tex]\sqrt{2}[/tex])
    or x = (2n+1)π - asin(1/[tex]\sqrt{2}[/tex])

    b) 2nπ ± acos(1/2)

    Just wondering,

    for a) how do they find the second solution x = (2n+1)π - asin(1/[tex]\sqrt{2}[/tex])? I'm curious to know, as the follow on question asks for the first 2 positive solutions to sin x = 1/[tex]\sqrt{2}[/tex], which can apparently only be obtained using both answers shown above. I only know how to get the first solution

    for b) where does the ± come from? My answer is the same but for ±.

    thanks in advance
     
  2. jcsd
  3. Sep 20, 2010 #2
    hmm interesting.
    got an email notification saying there was a reply to this thread, but there is nothing here!

    guess posted must have deleted!
     
  4. Sep 20, 2010 #3
    Have you sketched both of them and tried to reason by looking at the sketches? It should be quite helpful to do that.

    Some things to note; cos is an even function as you'll see when you sketch it which is related to the +-.

    Sin is an odd function, and therefore is unlikely to have just one solution for all, hopefully when you sketch it you'll see that.
     
  5. Sep 20, 2010 #4
    I understand graphically how the solutions are valid, but is there a set method/recipe to obtain these results?

    Would it be right ro assume, that for the cos question, my method is correct, it just needs the plus/minus in there?

    as for the sin question, how they get the second general solution is still a puzzle to me...
     
  6. Sep 20, 2010 #5

    HallsofIvy

    User Avatar
    Science Advisor

    Just adding multiples of 2pi is not enough (saying that sin(x) has period 2pi means that if y and x differ by a multiple of 2pi, then sin(x)= sin(y). It does NOT say that if sin(x)= sin(y) then they must differ by a multiple of 2pi.

    One way to remember the connections is to think of sine and cosine in terms of the unit circle. If (x, y) is a point on the unit circle so that the radius make angle [itex]\theta[/itex] with the positive x-axis, then [itex]x= cos(\theta)[/itex] and [itex]y= sin(\theta)[/itex]. That means that 2 angles, [itex]\theta[/itex] and [itex]\phi[/itex], between 0 and [itex]2\pi[/itex] will have the same sine if they lie on same horizontal line (same y value) and such a line will intersect the unit circle twice: by symmetry, [itex]\phi= \pi- \theta[/itex]. Similarly with cosine and a vertical line: [itex]cos(\theta)= cos(2\pi- \theta)[/itex].
     
  7. Sep 20, 2010 #6
    ok yep, definately see what you are saying (halls of ivy).

    when I look at sin, positive in 1st and 2nd quadrants, then when I look at the solutions it makes sense. same with cos, positive in 1st and 4th quadrants, the plus minus accounts for this.

    Naturally this leads to the following question:

    find general solution of:
    tanx = -3/[tex]\sqrt{3}[/tex]
    x = atan(-3/[tex]\sqrt{3}[/tex])

    tan negative in 2nd and fourth quadrants, so general equation(s) are
    (2n-1)π - atan(-3/[tex]\sqrt{3}[/tex])
    2nπ - atan(-3/[tex]\sqrt{3}[/tex])

    Sound good?
     
  8. Sep 20, 2010 #7
    bump.

    the negative in the atan(.....) is bothering me.
    I still need to be convinced!
     
  9. Sep 23, 2010 #8
    bump
    still seeking clarification with the negative case
     
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