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Trigonometry Question from a Math Contest

  1. Jan 5, 2005 #1
    sin^8(x) + cos^8(x) = 97 / 128
     
  2. jcsd
  3. Jan 5, 2005 #2

    Zurtex

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    The only way I can think off the top of my head is say it is equivalent to:

    [tex]128 \left[ \sin^8x + \left(1 - \sin^2x \right)^4 \right] - 97 = 0[/tex]

    And really hope it is solvable.
     
  4. Jan 5, 2005 #3
    I'd just keep pluggin in trig identities until you reduce things either all to single powers or maybe a quadratic equation of cosines. It shouldn't be too hard, you can use

    2cos^2(x) = 1+cos(2x)
    2sin^2(x) = 1 - cos(2x)
     
  5. Jan 5, 2005 #4

    krab

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    It would be hard to prove because it's not true.

    Edit: Sorry. I assumed you meant it as being true for all x. Do you it is to be solved for x?
     
    Last edited: Jan 5, 2005
  6. Jan 5, 2005 #5

    dextercioby

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    It can be reduced to two biquadratic equations,which are solvable.
    [tex] 1=[(\sin^{2}x+\cos^{2}x)^{2}]^{2}=(\sin^{4}x+\cos^{4}x+2\sin^{2}x\cos^{2}x)^{2}[/tex]
    [tex]=\sin^{8}x+\cos^{8}x+4\sin^{4}x \cos^{4}x+2\sin^{4}x \cos^{4}x+4\sin^{2}x \cos^{2}x (\cos^{4}x+\sin^{4}x) [/tex] (1)

    From (1) it follows:
    [tex] 1=\sin^{8}x+\cos^{8}x+6\sin^{4}x \cos^{4}x+4\sin^{2}x\cos^{2}x[(\sin^{2}x+\cos^{2}x)^{2}-2\sin^{2}x\cos^{2}x][/tex]
    [tex]=\sin^{8}x+\cos^{8}x-2\sin^{4}x \cos^{4}x+4\sin^{2}x\cos^{2}x[/tex] (2)

    From (2) it follows:
    [tex]\sin^{8}x+\cos^{8}x=1+2\sin^{4}x\cos^{4}x-4\sin^{2}x\cos^{2}x [/tex] (3)

    I use:
    [tex] \sin 2x=2\sin x\cos x [/tex](4)
    to write
    [tex]+2\sin^{4}x\cos^{4}x=+\frac{1}{8}(\sin 2x)^{4} [/tex] (5)
    [tex]-4\sin^{2}x\cos^{2}x=-(\sin 2x)^{2} [/tex] (6)

    Therefore,combining (5),(6) and (3),one gets:
    [tex]\sin^{8}x+\cos^{8}x=1+\frac{1}{8}(\sin 2x)^{4}-(\sin 2x)^{2} [/tex] (7)

    The initial equation
    [tex]\sin^{8}x+\cos^{8}x=\frac{97}{128}[/tex] (8)
    ,using (7),becomes
    [tex] 1+\frac{1}{8}(\sin 2x)^{4}-(\sin 2x)^{2}=\frac{97}{128} [/tex] (9)
    ,which can be put in the form
    [tex] \frac{1}{8}}(\sin 2x)^{4}-(\sin 2x)^{2}+\frac{31}{128}=0 [/tex] (10)

    Multiplying (10) by [itex] +8 [/itex] and making the obvious substitution
    [tex] (\sin 2x)\rightarrow y [/tex] (11)
    ,the equation (8) is brought to the canonical form of an biquadratic algebraic eq. in the variable "y":
    [tex] y^{4}-8y^{2}+\frac{31}{16}=0 [/tex](12)
    We make the substitution:[itex] y^{2}\rightarrow z [/itex] (13)
    ,and the eq.(12) becomes
    [tex] z^{2}-8z+\frac{31}{16}=0 [/tex] (14)
    ,with the solutions
    [tex]z_{1}=4+\frac{\sqrt{225}}{4}=\frac{1}{4}[/tex] (15)
    [tex]z_{2}=4-\frac{\sqrt{225}}{4}= \frac{31}{4}[/tex](16)

    Comparing (11),(13) and (16),we conclude that [itex] z_{2} [/itex] is not a viable solution.
    To be continued.
     
    Last edited: Jan 5, 2005
  7. Jan 5, 2005 #6

    dextercioby

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    From (13) and (15),we obtain:
    [tex] y^{2}=+\frac{1}{4} [/tex] (17)
    ,with the solutions:
    [tex] y_{1}=-\frac{1}{2} [/tex] (18)
    [tex] y_{2}=+\frac{1}{2} [/tex] (19)

    It can be easily checked that both solutions (18) and (19) are found in the interval (-1,+1),so they provide viable solutions to the problem.
    Putting together (18),(19) and (11),we found two transcendental equations
    [tex] \sin 2x=-\frac{1}{2}[/tex] (20)
    [tex] \sin 2x=+\frac{1}{2}[/tex] (21)

    We solve these eq.in the interval [itex] (-\frac{\pi}{2},+\frac{\pi}{2}) [/itex](for the 'sine' argument,which is 2x).In this case,we can apply the inverse function 'arcsine' to find
    [tex] x_{1}=\frac{1}{2}\arcsin(-\frac{1}{2})=-\frac{\pi}{12}[/tex] (22)
    [tex] x_{2}=\frac{1}{2}\arcsin(+\frac{1}{2})=+\frac{\pi}{12} [/tex] (23)

    Daniel.

    EDIT:Thank you,Krab,i wouldn't have seen it,years after...
     
    Last edited: Jan 5, 2005
  8. Jan 5, 2005 #7

    krab

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    Looks like it's [itex]\pi/12[/itex]
     
  9. Jan 5, 2005 #8

    dextercioby

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    Let's take [itex]\pi\sim 3.14 [/itex]
    [tex] \sin\frac{\pi}{12}\sim 0.25869 [/tex]

    [tex] \sin^{8}\frac{\pi}{12}\sim 2\cdot 10^{-5} [/tex](1)

    [tex] \cos\frac{\pi}{12}\sim 0.96596 [/tex]

    [tex] \cos^{8}\frac{\pi}{12}\sim 75800\cdot 10^{-5} [/tex] (2)

    Add (1) and (2) and get:
    [tex] \sin^{8}\frac{\pi}{12}+\cos^{8}\frac{\pi}{12}\sim 75802\cdot 10^{-5} \sim\frac{97}{128} [/tex] (3)


    Daniel.
     
    Last edited: Jan 5, 2005
  10. Jan 5, 2005 #9

    Hurkyl

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    Let s = sin^2 x, c = cos^2 x be a solution to the equation.

    As Zurtex Started, this equation is a quartic in s:

    97/128 = s^4 + (1-s)^4

    The symmetry suggests making the substitution s = t + 1/2

    Note that t is in the range -1/2 <= t <= 1/2, if we assume to look for real solutions.

    97/128 = (1/2 + t)^4 + (1/2 - t)^4
    = 2t^4 + 3t^2 + 1/8 = 2t^4 + 3t^2 + 16/128

    so

    2t^4 + 3t^2 - 81/128 = 0

    yielding the solution:

    t^2 = -3/4 +/- (1/4) sqrt(9 + 81/16)
    = -3/4 +/- (1/4) sqrt(225/16)
    = -3/4 +/- (1/4) (15/4)
    = -12/16 +/- 15/16
    = (-12 +/- 15) / 16

    t^2 = -27 / 16 or 3 / 16

    However, we know that 0 <= t^2 <= 9/4, so we have the single solution:

    t^2 = 3/16

    so

    t = sqrt(3) / 4

    s = t + 1/2 = (2 + sqrt(3)) / 4

    Now, s = sin^2 x, so we have:

    sin x = (1/2) sqrt(2 + sqrt(3))

    Where both square roots could be either positive or negative.



    For some reason, I feel that knowing a lot of solutions to the original equation should help me -- if x is a solution, then so is -x, ix, and 90 - x, from which you can make lots more solutions -- but I couldn't get such an approach to help.
     
    Last edited: Jan 5, 2005
  11. Jan 5, 2005 #10

    krab

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    But it's not meant to be 1; it's meant to be 97/128, which it is.

    BTW, I found your error. In your equation 1, your last 2 should be a 4. Apart from this, the technique is really nice.
     
  12. Jan 5, 2005 #11

    dextercioby

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    Hurkyl,i found two solutions in the reals,in the interval [itex] (-\frac{\pi}{4},+\frac{\pi}{4}) [/itex],because i found it easier to use 'arcsin'.
    Solving the eq.
    [tex] \sin 2x=\pm \frac{1}{2} [/tex]
    on the whole real exis would yield all real solutions to the equation.

    If u wanna inlcude complex solutions as well,u should use not circular,but hyperbolic trigonometry (identities and stuff).

    Daniel.
     
  13. Jan 6, 2005 #12

    Curious3141

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    The solutions have already been given, but perhaps the most "direct" way to do this would be like so :


    Let [itex]c = \cos \theta[/itex], [itex]s = \sin \theta[/itex], [itex]S = \sin 2\theta[/itex]

    [tex]s^8 + c^8 = 97/128[/tex]

    [tex](c^4 + s^4)^2 - 2(c^2s^2)^2 = \frac {97}{128}[/tex]

    [tex]((c^2 + s^2)^2 - 2c^2s^2)^2 - 2(\frac{1}{4}S^2)^2 = \frac {97}{128}[/tex]

    [tex](1 - \frac{1}{2}S^2)^2 - 2(\frac{1}{4}S^2)^2 = \frac {97}{128}[/tex]

    [tex](1 - \frac{1}{2}S^2)^2 - \frac{1}{8}S^4 = \frac {97}{128}[/tex]

    Expanding the LHS, rearranging and simplifying,

    [tex]\frac{1}{8}S^4 - S^2 + \frac{31}{128} = 0[/tex]

    which is a quadratic equation in S^2. Solving that will give the required general solution in the reals :

    [tex]\theta = n\pi (+/-) \frac{\pi}{12} [/tex] or [tex]n\pi (+/-) \frac{5\pi}{12}[/tex]
     
    Last edited: Jan 6, 2005
  14. Jan 7, 2005 #13
    oh boy, anyone in highschool taking the amc 12 in a few weeks?
     
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