Challenge 82: Cosine to the 2019th Power

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  • Published on Feb 28, 2019
  • Congratulations to attyfarbuckle, alonamaloh, Rishav Gupta, Allaizn, Nicola C, Jaleb, and Sonal Kumari for successfully solving the last week's math challenge question! attyfarbuckle was the first person to solve the question.
    Welcome, everyone! LetsSolveMathProblems presents one weekly math challenge problem every Wednesday (continental U.S. time). To participate, please comment your proposed answer and explanation below (or discuss potential solutions with other viewers), keeping in mind that your comment should be left unedited to win the recognition prize. Up to the first ten people with correct solutions are recognized in the next challenge video. The solution to a challenge problem is posted as a separate video the following Wednesday.
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Comments • 49

  • Ee Poh Ong
    Ee Poh Ong 20 days ago

    Yay I did something and got 0.132928

  • -_-
    -_- 3 months ago

    2692
    Ez pz

  • Bintang Alam Semesta W.A.M

    Solution:
    first, let's evaluate each term on the LHS:
    1+cot^4(x) = 1+(cos^4(x)/sin^4(x))
    csc^2(x)+tan^2(x) = (1/sin^2(x))+(sin^2(x)+cos^2(x)) = (sin^4(x)+cos^2(x))/(sin^2(x)cos^2(x))
    (tan(x/2))^2 = (sin^2(x))/(1+2cos(x)+cos^2(x))
    Second, let's remove (tan(x/2))^2 term from LHS to RHS then we'll have
    (sin^4(x)+cos^4(x))(sin^4(x)+cos^2(x)) = sin^4(x)cos^2(x)(1+2cos(x)+cos^2(x))
    after simplifying the above equation, we'll obtain
    sin^8(x) + cos^6(x) = 2sin^4(x)cos^3(x) ...........(*)
    let's apply variabel manipulation for sin^4(x) and cos^2(x) respectively,
    q = sin^4(x) ; r= cos^2(x)
    (substituting these into ......(*))
    then,
    q^2+r^3 = 2qr^(3/2) (q^2+r^3)^2 = 4q^2r^3 q^4-2q^2r^3+r^6 = 0 (q^2-r^3)^2 = 0.
    then we'll get q^2=r^3
    sin^8(x) = cos^6(x)
    cos(x) = (sin(x))^(4/3) ......(**)
    thus,
    cos^2019(x) = (sin(x))^((4/3)*2019) = sin^2692(x),
    Hence,
    k=2692.

  • Piero Puelles
    Piero Puelles 3 months ago

    2019

  • Intergalactic Bajrang dal

    tan(x/2) can be writeen as sqrt((1-cosx)/(1+cosx)). writing the other tanx, cosecx,cotx in terms of sinx and cosx and then expanding the LHS, shifting the denominator on the RHS we can the expression:
    cos^6(x) + cos^2(x)sin^4(x)+cos^4(x)sin^4(x) +sin^8(x)-cos^7(x)-cos^3(x)sin^4(x)-cos^5(x)sin^4(x)-cosxsin^8(x) = sin^6(x)cos^2(x)+sin6^(x)cos^3(x)-cos^4(x)sin^4(x)
    using the identity cos^2(x) +sin^2(x) = 1 multiple times we finally get one of the solutions as
    cos^3(x)= sin^4(x)
    raising both sides to the 673rd power we get
    cos^2019(x)= sin^2692(x)
    k=2692

  • Acute angle
    Acute angle 3 months ago

    1. I woke up.
    2. Check my phone.
    3. Got a notification from USclip.
    4. My comment has been chosen!
    Actually the correct sequence: 2>3>4>1😭

  • UbuntuLinux
    UbuntuLinux 3 months ago

    Following this: www.slideshare.net/mobile/nguyenhoanhtnut/nhung-cong-thuc-luong-giac-co-ban (although this is not in Eng but it has the formulas)
    and goo.gl/images/R7zKcm
    tan^2(x/2) = sin^2(x)/(1+cos(x))^2
    cot^4(x)+1 = cos^4(x)/sin^4(x) + 1 = [cos^4(x)+sin^4(x)]/sin^4(x)
    csc^2(x)+tan^2(x)=1/sin^2(x)+sin^2(x)/cos^2(x)
    Replacing all in we get
    s^2(c^4+s^4)(c^2+s^4)=(1+c)^2*s^4*c^2*s^2
    with s = sinx, c = cosx
    Which simpifies itself to
    (c^4+s^4)(c^2+s^4)=(1+c^2)^2*s^4*c^2= s^4*c^2*(1+2c+c^2)
    c^6 + c^4*s^4 + s^4*c^2 + s^8 = s^4*c^2 + 2s^4*c^3 + s^4*c^4
    This leads to c^6 + s^8 = 2s^4*c^3, or (c^3-s^4)^2=0, or c^3 = s^4
    Hence, (cos^3(x))^673=(sin^4(x))^673
    or cos^2019(x) = sin^2692(x)

  • Davide
    Davide 3 months ago

    expressing everything in terms of sin and cos we get sin^4(x)=cos^3(x), implying sin^2692(x)=cos^2019(x) thus the answer is 2692

  • Fernando Peñaherrera
    Fernando Peñaherrera 3 months ago

    k=2019*4/3. The above identity simplifies to (cos(x))^3=(sin(x))^4

  • Shiv Bhardwaj
    Shiv Bhardwaj 3 months ago

    K =2692

  • 정태우
    정태우 3 months ago

    I cant understand how they solved this :( Did anybody can follow their explanations??
    I really wonder how everybody get 2692.

  • T Rat #Math-info
    T Rat #Math-info 3 months ago

    I don't know what csc is :'(

  • Mohamed Sheriff
    Mohamed Sheriff 3 months ago

    Rewrite the equation in terms of Sin x and Cos x. Now,let u = Sin x and v = Cos x. Substitute these results into the equation and reduce the equation to :
    v³ = u⁴
    ⇒ v = u^4/3 ----(1)
    From :
    Cos^2019(x) = Sin^k(x)
    It follows that :
    v^(2019) = uᵏ ----(2)
    Now,substitute the result of v in equation into equation (2).
    (u^4/3)²⁰¹⁹ = uᵏ
    ∴ u²⁶⁹² = uᵏ
    Hence,equate the powers since the bases are equal.
    ∴ k = 2692

  • EyyLmaoo
    EyyLmaoo 3 months ago +1

    Simplify first bracket as (1-cosx)/(1+cos) using cosine double angle identity subbing in x/2 for theta.
    Translate the cot, csc, and tan into sines and cosines.
    Simplify using identities
    Left with (cos^3(x)-sin^4(x))^2=0
    Taking the square root, cos^3(x)-sin^4(x)=0
    cos^3(x)=sin^4(x)
    Take to the power of 2019/3 or 673:
    cos^2019(x)=sin^(673*4)(x)=sin^(k)(x)
    Therefore k=673*4= 2692

  • adandap
    adandap 3 months ago

    I suspect I am way late to the party, but here is my solution.
    First, we need to get rid of the x/2, so we write tan(x/2) = sin(x)/(1+cos(x)) and multiply out the given expression so there are no fractions. Using the notation c = cos(x), s=sin(x) we get c^2 (1+c^2) s^4 = (c^4 + s^4) (c^2 +s^4). Happily, this is quadratic in s^4, and in fact reduces to (s^4-c^3)^2 = 0. So cos^3(x) = sin^4(x). Noting that 2019 = 3 * 673, then k = 673 *4/3 = 2692.

  • mstmar
    mstmar 3 months ago +1

    k = 2692
    if we replace tan(x/2) with its half angle formula (sin/(1+cos), and write cot, csc and tan in terms of sin and cos, then some slight simplifications, we get
    1/(1+cos(x))^2 * (cos(x)^4+sin(x)^4)/sin(x)^4 * (cos(x)^2 + sin(x)^4)/cos(x)^2 = 1
    multiply both sides by (1+cos(x))^2 * sin(x)^4 * cos(x)^2, and subtract right from left
    cos(x)^6+sin(x)^8 - 2*cos(x)^3*sin(x)^4 = 0
    this factors to
    (cos(x)^3-sin(x)^4)^2
    thus
    cos(x)^3=sin(x)^4
    raising both sides to the 673rd power, we get
    cos(x)^2019=sin(x)^2692

  • Carlo Eduardo Chambi Torrico

    x=47.855°

  • Serengeti Ghasa
    Serengeti Ghasa 3 months ago

    The answer is 2692
    I found as follows
    tan²x/2.(cos⁴x+1)(csc²x+tan²x)
    ={sin²x/(1+cosx)²}×{(cos⁴x+sin⁴x)/sin⁴x}×{(cos²x+sin⁴x)/sin²x.cos²x}=1
    then simplify i got
    cos⁶x+sin⁸x=2cos³x.sin⁴x
    => cos³x=sin⁴x
    =>(cos³x)⁶⁷³=(sin⁴x)⁶⁷³
    =>cos²⁰¹⁹x=sin²⁶⁹²x
    that is k=2692
    with love
    Prabhat Kumar Sahu

  • Keith
    Keith 3 months ago

    tan(x/2)=sin(x)/ (1+cos(x)) so
    tan^2(x) = sin^2(x)/ (1+cos(x))^2
    cot^4(x) +1 = cos^4(x)/sin^4(x) +1 = [cos^4(x) +sin^4(x)]/ { sin^4(x)}
    csc^2(x)+tan^2(x) = 1/sin^2(x) + sin^2(x)/cos^2(x) = [cos^2(x)+sin^4(x)] / {cos^2(x) sin^2(x)}

    Then when we move the denominators to the right side
    sin^2(x) [cos^4(x) +sin^4(x)] [cos^2(x)+sin^4(x)] = (1+cos(x))^2) { sin^4(x)} {cos^2(x) sin^2(x)}

    divide through by sin^2(x) then expand and reduce
    cos^6(x) +cos^2(x) sin^4(x) +cos^4(x) sin^4(x) +sin^8(x) = cos^2(x) sin^4(x) +2cos^3(x)sin^4(x) +cos^4(x)sin^4(x)
    cos^6(x)+sin^8(x) = 2cos^3(x)sin^4(x)

    cos^6(x) - 2 cos^3(x) sin^4(x) + sin^8(x) =0
    (cos^3(x) -sin^4(x))^2=0

    Simplifies to sin^4(x)=cos^3(x) and 2019/3 =673,
    Thus (cos^3(x))^673= (sin^4(x))^673, k = 4*673 = 2692

    • LetsSolveMathProblems
      LetsSolveMathProblems  3 months ago

      As of now, your solution is not quite comprehensive enough to be considered for recognition. Could you mention some of the trigonometric identities you used or algebraic manipulations you performed to obtain sin^4(x) = cos^3(x)?
      * Please do not edit your comment, but instead post a new comment with the modified solution.

  • divyansh mittal
    divyansh mittal 3 months ago

    Easy 2692 , Simplify as sinx and cosx ...

  • Hiren Bavaskar
    Hiren Bavaskar 3 months ago

    My answer is 2692
    First convert everything into cos and sines...
    Tan^ 2(x/2)= (1-cos x) /(1+cosx)
    Now expand everything. Finally for symmetry multiply 1+cos x on both sides to get (1-cos x)(1+ cos x)(..rest terms) = (1+ cosx )^2 *sin^6 x* cos^2 x....
    Now 1- cos^2 x is sin^2 x and now expand and cancel...
    We get (sin^4 x- cos^3 x)^2= 0
    Hence sin^4 x= cos^3 x
    Now 2019=673*3
    So raise both sides to 673 ..
    Hence k= 2692

    • Hiren Bavaskar
      Hiren Bavaskar 3 months ago

      I would like to add something I haven't written... On multiplying (1+cosx ) to both sides.. we finally gets sin^2 x * ( sin^4 x- cos^3 x) = 0.. obviously here sin^2 x= 0 is a false root as it doesn't satisfy the original equation..
      So we get (sin^4 x - cos^3 x)^2=0
      And we finally get to value of k

  • mark erena
    mark erena 3 months ago

    The awnser is 2692.
    Writing the expresion in terms of sin x and cos x and symplifing this you get (sin^4 x - cos^3 x)^2=0 That means sin^4 x - cos^3 x =0 or sin^4 x = cos^3 x. That means that if you raise it to the 673th power you get sin^2692 x= cos^2019 x Thus 2692

    • LetsSolveMathProblems
      LetsSolveMathProblems  3 months ago

      As of now, your solution is not quite comprehensive enough to be considered for recognition. Could you mention some of the trigonometric identities you used or algebraic manipulations you performed to obtain (sin^4x - cos^3x)^2=0?
      * Please do not edit your comment, but instead post a new comment with the modified solution.

  • Nitro Zox
    Nitro Zox 3 months ago

    2692
    Converting the given expression to sinx and cosx and tan^2(x/2) is 1-cosx/1+cosx, csc^2 + tan^2 is just cos^2 + sin^4/(sin^2x)(cos^2x) , we get cos^3x = sin^4x.
    Taking ln : lncosx/lnsinx = 4/3.
    Hence k/2019 = 4/3 . So k = 2692

    • Hiren Bavaskar
      Hiren Bavaskar 3 months ago

      csc^2 x + tan^2 x is not equal to 1/(sin^2 x * cos^2 x ) bro! If you used this.. how did you arrive at your final answer correctly?

  • JIN ZHI PHOONG
    JIN ZHI PHOONG 3 months ago

    Firstly, let s = sinx and c = cosx for simplicity.
    Then we have
    [tan(x/2)]^2 = [s/(1+c)]^2
    (cotx)^4 + 1 = (c^4+s^4)/s^4
    (cscx)^2 + (tanx)^2 = (c^2 + s^4)/(s^2*c^2)

    Putting them together we have:
    [s/(1+c)]^2 * (c^4+s^4)/s^4 * (c^2 + s^4)/(s^2*c^2) = 1
    Then moves all the denominators to the other side,
    s^2(c^4+s^4)(c^2+s^4) = s^6c^2(1+c)^2
    Expanding all the parenthesis we get,
    s^2c^6 + s^6c^2 + s^6c^4 + s^10 = s^6c^4 + 2s^6c^3 + s^6c^2
    s^2c^6 + s^10 = 2s^6c^3
    c^6 + s^8 = 2s^4c^3
    s^8 - 2s^4c^3 + c^6 =0
    (s^4 - c^3)^2 = 0
    s^4 - c^3 = 0
    therefore c^3 = s^4.
    Hence c^2019 = s^2692.

    k = 2692

  • Rishav Gupta
    Rishav Gupta 3 months ago +15

    The answer is 2692
    Simplyfying whole expression
    tan^2(x/2) can be written as
    sin^2(x/2)/cos^2(x/2)
    =(1-cosx)/(1+cosx)
    And simplyfing in the expression in terms of sines and cosines we get
    (cos^3(x)-sin^4(x))^2=0
    Which implies
    cos^3(x)=sin^4(x)
    Thus raising them to 673rd power we get
    cos^2019(x)=sin^2692(x)

    • Meet Patel
      Meet Patel 3 months ago

      @Rishav Gupta ya. The same goes with me. Cracked pre rmo but missed rmo with 4 marks behind cutoff. Nevertheless, I am quite sure about NTSE stage 2 selection

    • Rishav Gupta
      Rishav Gupta 3 months ago

      @Arnav kumar Sinha I am from bihar and my ntse stage 1 results will come in mid march

    • Arnav kumar Sinha
      Arnav kumar Sinha 3 months ago +1

      @Rishav Gupta what is ur rank in NTSE?

    • Arnav kumar Sinha
      Arnav kumar Sinha 3 months ago +1

      @Rishav Gupta wow bro prmo is also a great thing .from which state u are?

    • Rishav Gupta
      Rishav Gupta 3 months ago

      @NILESH KUMAR I have just cleared prmo but not rmo.

  • linkzaum
    linkzaum 3 months ago

    got x being 0.8352ish with wolframalpha
    so k would be 2692 (2691.99, but i used the approximated x value)

    • LetsSolveMathProblems
      LetsSolveMathProblems  3 months ago +1

      It is more than fine. I thought I should promptly let you know that your answer would not be counted unless it could be reproduced without a computer. Quite frankly, I encourage using programs like WolframAlpha to advance your understanding of mathematics. =)

    • linkzaum
      linkzaum 3 months ago

      @LetsSolveMathProblems oh sorry about this, you can remove my comment if you want, i only used a external program because im kinda of slow with math and i wanted to be in the first ten

    • LetsSolveMathProblems
      LetsSolveMathProblems  3 months ago

      For this particular problem, solutions that rely on computer programs will not be accepted. Of course, computer programs may be used to investigate the problem or guess the final answer, but the final solution that is posted should be independent of such programs. (The exceptions are enumerative combinatorics questions, for which I usually accept solutions via coding.)

  • JHawk24
    JHawk24 3 months ago +3

    My answer is 2692

    By graphing it on demos, I found that x is about 0.8352.
    If (cos(x))^2019 = (sin(x))^k
    then log base sin(x) of (cos(x))^2019 = k,
    reforming this,
    2019ln(cos(x))/ln(sin(x)) = k
    plugging in the approximation 0.8352 gives that k is about 2692.

    • LetsSolveMathProblems
      LetsSolveMathProblems  3 months ago +5

      For this particular problem, solutions that rely on computer programs will not be accepted. Of course, computer programs may be used to investigate the problem or guess the final answer, but the final solution that is posted should be independent of such programs. (The exceptions are enumerative combinatorics questions, for which I usually accept solutions via coding.)

  • replicaacilper
    replicaacilper 3 months ago +1

    What the hell

  • George Pi Full
    George Pi Full 3 months ago

    aha