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Trigonometric Equations

  1. Aug 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the solution set of the equation below for the interval 0° ≤ x ≤ 360°:

    arccos(x) - arcsin(1-x) = 90°

    2. Relevant equations
    all given above.


    3. The attempt at a solution
    i tried to consider (1-x) as "complementary" but I'm not really sure about that and I would like to know your opinion if I should go for it? And I'm kinda confused why the RHS is in degrees. Is it because the problem is dealing with inverse trig equations (which the answers are supposed to be angles)? Thanks in advance.
     
  2. jcsd
  3. Aug 29, 2012 #2

    Simon Bridge

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    That doesn't make sense - surely it is -1 ≤ x ≤ 1 (angle is the output of inverse trig functions).

    Hmmm ... well, [tex]\arccos(x)+\arcsin(x)=90^\circ[/tex]

    We-ell, there are a whole lot of trig identities that may be useful too.
    ... the RHS is the result of the sum of two angles, so, naturally it is in degrees.

    I'd have been tempted to take the cosine of both sides and use the sum-to-product relations with the trig-inverse-trig relations to get some f(x)=0 ... then it's a matter of finding the roots of f.

    But your idea is cool too ... if you can show that arcsin(1-x) = -arcsin(x) you'd be made.

    eg. if x=1, then 1-x=0, arcsin(1)=90, arcsin(0)=0 or 180.
    So - not generally true. But maybe it is true for some value of x?

    Of course, plotting the function y=arccos(x)-arcsin(1-x) could give a few clues.
     
  4. Aug 29, 2012 #3

    micromass

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    No particular reason, I think. You can write them in radians too if you wish. So you could as well write

    [tex]arccos(x)-arcsin(1-x)=\frac{\pi}{2}[/tex]

    It's the same thing, although I prefer to write them like this.

    Anyway, try to write the equation as

    [tex]arccos(x)=\frac{\pi}{2}+arcsin(1-x)[/tex]

    and now take the cosine of both sides.
     
  5. Aug 29, 2012 #4

    Hi, micromass. Thanks for posting a reply.
    I'm working on it now and I just want to confirm if I'm doing this right:

    Taking cosine of both sides will give

    [tex]0 = \cos\frac{\pi}{2} + \cos[\arcsin(1-x)][/tex]

    Question: In cos[arcsin(1-x)], is it right to use "substitution" method where I let, say θ, equal to arcsin(1-x), then find for cosθ afterwards?

    It's something like: cos[arcsin(1-x)]

    let θ = arcsin(1-x)

    sinθ = 1-x then is it right to find for cosθ through this?

    Thanks!

    EDIT: I guess this is wrong T_T Please tell me and sorry coz I'm still a learner T_T
    EDIT: I'm having trouble with latex again.
     
    Last edited: Aug 29, 2012
  6. Aug 29, 2012 #5

    Mentallic

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    [tex]\cos(A+B)\neq \cos(A)+\cos(B)[/tex]

    Remember it's

    [tex]\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)[/tex]

    Yes you're on the right track. So if [itex]\sin(\theta) = 1-x[/itex] then draw up a right-angled triangle with one angle being [itex]\theta[/itex] and then label the sides in such a way such that [itex]\sin(\theta)=1-x[/itex]
    Then, since we are actually looking for the value of [itex]\cos(\arcsin(1-x))\equiv \cos(\theta)[/itex] you'll need to find the value of the other side in the right-triangle.
     
  7. Aug 29, 2012 #6
    Oh yes!! Why do I always forget that. Thanks a lot! Ok, I'm gonna work on it again. Thank you for all your replies! :)
     
  8. Aug 29, 2012 #7
    Alright. Summing up everyone's ideas, I got this solution:

    [tex] cos[arccos(x) = \frac{\pi}{2} + arcsin(1-x)] [/tex]
    [tex] 0 = cos[\frac{\pi}{2} + arcsin(1-x)] [/tex]


    [tex]0 = cos\frac{\pi}{2}cos[arcsin(1-x)] - sin\frac{\pi}{2}cos[arcsin(1-x)[/tex]
    [tex]0 = cos[arcsin(1-x)][/tex]

    [tex] let θ = arcsin(1-x)[/tex] ⇔ sinθ = 1-x

    for angle θ: y = 1-x, x = [itex]\sqrt{2x-x^2}[/itex], r = 1

    ∴ cosθ = [itex]\sqrt{2x-x^2}[/itex]

    [itex] 0 = -cos(θ) [/itex]
    [itex]0 = -(\sqrt{2x-x^2})[/itex]
    [itex]0 = \sqrt{2x-x^2}[/itex]
    [tex]0 = 2x-x^2[/tex]
    [tex]0 = (2-x)x[/tex]

    [tex]x = 2 , x = 0[/tex]

    so x = 0.

    Hope it's right now.
     
  9. Aug 29, 2012 #8

    Simon Bridge

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    note: arccos is an inverse cosine so:[tex]\cos(\arccos(x))=x[/tex]

    Also, you'll find there is an identity for [itex]\cos(\frac{\pi}{2}+\theta)[/itex] that will be useful for simplifying things.
    You'll still get it the long way around though.
     
  10. Aug 29, 2012 #9
    oh yeaahh. i forgot T____T
    how am i going to pass my test if i always keep on making mistakes.
    my test is already tomorrow, and i got that problem above from a sample problem of what might appear as question for tomorrow's exam.
    i think i would fail again :(
    how did you guys became so gifted in math?
     
  11. Aug 29, 2012 #10
    it's already late here and i'm still not yet through studying for exam :( please check my new solution. I still hope I could get it:


    [itex] x = -cos(θ) [/itex]
    [itex] x= -\sqrt{2x-x^2}[/itex]
    [itex] -x = \sqrt{2x-x^2}[/itex]
    [tex] -x^2 = 2x-x^2[/tex]
    [tex] 0 - 2x = 0[/tex]
    [tex] x = 0 [/tex]
     
  12. Aug 29, 2012 #11

    Mark44

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    When you square -x, you get x2, not -x2.
    You could have squared both sides of the 2nd equation, above.
     
  13. Aug 29, 2012 #12
    how about this:

    [itex] x = -cos(θ) [/itex]
    [itex] x= -\sqrt{2x-x^2}[/itex]
    [tex] x^2 = 2x-x^2[/tex]
    [tex] 2x^2 - 2x = 0[/tex]
    [tex] x^2 - x = 0 [/tex]

    x = 1, x = 0
     
  14. Aug 29, 2012 #13

    Mark44

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    That's better, but you're still not done. When you square both sides of an equation, there is the possibility that you are introducing extraneous solutions.

    Check both solutions to see if they satisfy the equation you squared: x = -√(2x - x2).
     
  15. Aug 29, 2012 #14
    The answer is x = 0.
    Because if x = 1, then

    [tex]x = -\sqrt{2x - x^2}[/tex]
    [tex](1) = -\sqrt{2(1) - 1^2}[/tex]
    [tex] 1 = -\sqrt{1}[/tex]
    [tex] 1 ≠ -1 [/tex]

    if x = 0

    [tex]0 = -\sqrt{0 - 0^2}[/tex]
    [tex](0) = -\sqrt{0}[/tex]
    [tex] 0 = 0 [/tex]

    Am I right sir?
     
  16. Aug 29, 2012 #15

    Mark44

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    That is the correct solution of the equation x = -√(2x - x2), but it's not a solution of your original equation. I haven't checked all of your work, but there must be an error somewhere.
     
  17. Aug 29, 2012 #16

    Simon Bridge

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    Yeh - not seeing where [itex]x=-\sqrt{2x-x^2}[/itex] comes from.

    [edit] no I got it - it's from the combined trig and inverse-trig identity for cos(arcsin(x))
    But that doesn't account for the extra [itex]\frac{\pi}{2}[/itex] in the RHS.

    He seems to have gone from:
    [tex]\arccos(x) = \frac{\pi}{2} + \arcsin(1-x)[/tex]... to
    [tex]x = \cos \big [ \arcsin(1-x) \big ] = \sqrt{1-(1-x)^2} [/tex] (don't know where the minus sign came from either.)

    I detect tiredness errors.

    hint: [itex]\cos(\frac{\pi}{2})=0[/itex]
     
    Last edited: Aug 29, 2012
  18. Aug 30, 2012 #17
    That was from extracting (1-x)^2
     
  19. Aug 30, 2012 #18

    Mentallic

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    [tex]\sqrt{x^2-2x}\neq - \sqrt{2x-x^2}[/tex]

    Remember the rule that [itex]\sqrt{ab}=\sqrt{a}\sqrt{b}[/itex] so this would imply that [itex]\sqrt{x^2-2x}=\sqrt{-(2x-x^2)}=\sqrt{-1}\sqrt{2x-x^2}[/itex] and the square root of -1 is not -1, it's imaginary. But the rule only applies for positive a and b values, so even then we can't do that.

    You need to leave it as it is.
     
  20. Aug 30, 2012 #19

    Simon Bridge

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    The minus sign outside the surd? Mentallic has described why that should not come from anything inside the surd.

    go back to:
    [tex]x=\cos\left [ \frac{\pi}{2} + \arcsin(1-x) \right ] [/tex]... put [itex]\theta=\arcsin(1-x)[/itex] and simplify: what does [itex]\cos(\frac{\pi}{2}+\theta)[/itex] turn into?
     
  21. Aug 30, 2012 #20
    this?

    let θ = arcsin(1-x)

    [tex]x=\cos\left [ \frac{\pi}{2} + \arcsin(1-x) \right ][/tex]

    [tex]x=\cos\left [ \frac{\pi}{2} + θ \right ][/tex]

    [tex]x=cos\frac{\pi}{2}cosθ - sinθsin\frac{\pi}{2}[/tex]

    x = -sinθ
     
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