# 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 3 months ago

Yay I did something and got 0.132928

• -_- 6 months ago

2692
Ez pz

• 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 6 months ago

2019

• 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 6 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 6 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 6 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 6 months ago

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

• Shiv Bhardwaj 6 months ago

K =2692

• 정태우 6 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 6 months ago

I don't know what csc is :'(

• Serengeti Ghasa 6 months ago

@T Rat #Math-info
U r welcome sir

• T Rat #Math-info 6 months ago

@Serengeti Ghasa oh ! Big thank to you sir

• Serengeti Ghasa 6 months ago

Hi Sir, CSC stand for cosec =1/sin
With love

• Mohamed Sheriff 6 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 6 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 6 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 6 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

• x=47.855°

• Serengeti Ghasa 6 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 6 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  6 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 6 months ago

Easy 2692 , Simplify as sinx and cosx ...

• Hiren Bavaskar 6 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 6 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 6 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  6 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 6 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 6 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 6 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 6 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 6 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 6 months ago

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

• Arnav kumar Sinha 6 months ago +1

@Rishav Gupta what is ur rank in NTSE?

• Arnav kumar Sinha 6 months ago +1

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

• Rishav Gupta 6 months ago

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

• linkzaum 6 months ago

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

• LetsSolveMathProblems  6 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 6 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  6 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 6 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  6 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 6 months ago +1

What the hell

• George Pi Full 6 months ago

aha