# 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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Ee Poh Ong20 days agoYay I did something and got 0.132928

-_-3 months ago2692

Ez pz

Bintang Alam Semesta W.A.M3 months agoSolution:

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 Puelles3 months ago2019

Intergalactic Bajrang dal3 months agotan(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 angle3 months ago1. 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>1š

UbuntuLinux3 months agoFollowing 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)

Davide3 months agoexpressing 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 PeĆ±aherrera3 months agok=2019*4/3. The above identity simplifies to (cos(x))^3=(sin(x))^4

Shiv Bhardwaj3 months agoK =2692

ģ ķģ°3 months agoI cant understand how they solved this :( Did anybody can follow their explanations??

I really wonder how everybody get 2692.

T Rat #Math-info3 months agoI don't know what csc is :'(

Serengeti Ghasa3 months ago@T Rat #Math-info

U r welcome sir

T Rat #Math-info3 months ago@Serengeti Ghasa oh ! Big thank to you sir

Serengeti Ghasa3 months agoHi Sir, CSC stand for cosec =1/sin

With love

Mohamed Sheriff3 months agoRewrite 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 :

vĀ³ = uā“

ā v = u^4/3 ----(1)

From :

Cos^2019(x) = Sin^k(x)

It follows that :

v^(2019) = uįµ ----(2)

Now,substitute the result of v in equation into equation (2).

(u^4/3)Ā²ā°Ā¹ā¹ = uįµ

ā“ uĀ²ā¶ā¹Ā² = uįµ

Hence,equate the powers since the bases are equal.

ā“ k = 2692

EyyLmaoo3 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

adandap3 months agoI 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.

mstmar3 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 Torrico3 months agox=47.855Ā°

Serengeti Ghasa3 months agoThe answer is 2692

I found as follows

tanĀ²x/2.(cosā“x+1)(cscĀ²x+tanĀ²x)

={sinĀ²x/(1+cosx)Ā²}Ć{(cosā“x+sinā“x)/sinā“x}Ć{(cosĀ²x+sinā“x)/sinĀ²x.cosĀ²x}=1

then simplify i got

cosā¶x+sināøx=2cosĀ³x.sinā“x

=> cosĀ³x=sinā“x

=>(cosĀ³x)ā¶ā·Ā³=(sinā“x)ā¶ā·Ā³

=>cosĀ²ā°Ā¹ā¹x=sinĀ²ā¶ā¹Ā²x

that is k=2692

with love

Prabhat Kumar Sahu

Keith3 months agotan(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

LetsSolveMathProblems3 months agoAs 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 mittal3 months agoEasy 2692 , Simplify as sinx and cosx ...

Hiren Bavaskar3 months agoMy 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 Bavaskar3 months agoI 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 erena3 months agoThe 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

LetsSolveMathProblems3 months agoAs 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 Zox3 months ago2692

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 Bavaskar3 months agocsc^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 PHOONG3 months agoFirstly, 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 Gupta3 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 Patel3 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 Gupta3 months ago@Arnav kumar Sinha I am from bihar and my ntse stage 1 results will come in mid march

Arnav kumar Sinha3 months ago^{+1}@Rishav Gupta what is ur rank in NTSE?

Arnav kumar Sinha3 months ago^{+1}@Rishav Gupta wow bro prmo is also a great thing .from which state u are?

Rishav Gupta3 months ago@NILESH KUMAR I have just cleared prmo but not rmo.

linkzaum3 months agogot x being 0.8352ish with wolframalpha

so k would be 2692 (2691.99, but i used the approximated x value)

LetsSolveMathProblems3 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. =)

linkzaum3 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

LetsSolveMathProblems3 months agoFor 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.)

JHawk243 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.

LetsSolveMathProblems3 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.)

replicaacilper3 months ago^{+1}What the hell

George Pi Full3 months agoaha