# Interesting Math Challange

If $\displaystyle{{ f }_{ n }(x)={ \cos ^{ n }{ x } +\cos ^{ n }{ (x+\cfrac { 2\pi }{ 3 } ) } +\cos ^{ n }{ (x+\cfrac { 4\pi }{ 3 } ) } }}$. Then Solve for x if $\displaystyle{{ f }_{ 7 }(x)=0}$.

Use Any Tool of Mathematics. Can You Guess why I choose $\displaystyle{\cfrac { 2\pi }{ 3 } \& \cfrac { 4\pi }{ 3 } }$. ? Note by Deepanshu Gupta
5 years ago

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$y = cosx + isinx$

$\displaystyle{y^{ n }+\cfrac { 1 }{ y^{ n } } =2cosnx=(cosx+isinx)^{ n }=cosnx+isinnx}$.

$y + \dfrac{1}{y} = 2cosx$

$cos^{7}x = ( y + \dfrac{1}{y})^{7}$

$= y^{7} + 7y^{5} + 21y^{3} + 35y + 35\dfrac{1}{y} + 21\dfrac{1}{y^{3}} + 7\dfrac{1}{y^{5}} + \dfrac{1}{y^{7}}$

$= ( y^{7} + \dfrac{1}{y^{7}}) + 7(y^{5} + \dfrac{1}{y^{5}}) + 21(y^{3} + \dfrac{1}{y^{3}}) + 35(y + \dfrac{1}{y})$

$= 2cos7x + 7.2cos5x + 21.2cos3x + 35.2cosx$

$cos^{7}x = \dfrac{1}{64}(cos7x + 7cos5x + 21cos3x + 35cosx)$

$cos^7(x+ \dfrac{2\pi}{3}) = \dfrac{1}{64}( - cos(7x - \dfrac{\pi}{3}) - 7cos(5x + \dfrac{\pi}{3}) + 21cos3x - 35cos(x - \dfrac{\pi}{3}))$

$cos^7(x + \dfrac{4\pi}{3}) = \dfrac{1}{64}( - cos(7x + \dfrac{\pi}{3}) - 7cos(5x - \dfrac{\pi}{3}) + 21cos3x - 35cos(x + \dfrac{\pi}{3}))$

$cosA + cosB = 2cos(\dfrac{A + B}{2})cos(\dfrac{A - B}{2})$

$cos^7(x+ \dfrac{2\pi}{3}) + ( cos^7(x + \dfrac{4\pi}{3}) = \dfrac{1}{64}( - cos7x - 7cos5x + 42cos3x - 35cosx)$

$Expression = \dfrac{63}{64}cos3x = 0 =cos\dfrac{\pi}{2}$

$3x = n\pi \pm \dfrac{\pi}{2}$

$x = \dfrac{n\pi}{3} - \dfrac{\pi}{6}$

- 5 years ago

Nice Megh , I'm Expecting It from You :) . But Still There is one more way of similar Type ! Just Focus on the the angles given in question .I beleive that You will Definitely Get it , just give one more Try !$\ddot\smile$

- 5 years ago

why Complex Numbers Tag ? :O

- 5 years ago

See megh solution , you will find it interesting :)

- 5 years ago

do we have to use binomial expanision In the mid of the solution . plz ans my question............ and reply

- 5 years ago

Yes We Have To use Binomial , But Don't Need to write whole Terms , By Sharp Observation we see that it will cancel out , You Have To Check only Those Terms which are independent of w and w^2 (By using General term for binomial)

- 5 years ago

$(e^{ix} + e^{-ix})/2 = cosx$ given equatin = 0 this implies, $((e^{ix} + e^{-ix})/2)^{7} (1+w^2 +w) =0$ this implies for all x......

- 5 years ago

Great ! You Catch Right Path , But You r doing calculation mistake , Infact You can't able To take common $(1+w^2 +w)$ Check ur calculation carefully that wheather you are writing $w$ or $w^2$.

- 5 years ago

its a long question do we have to apply binomial theorem(to expand) . Actuslly I got the answer which megh did. GREAT QUESTION. WHERE DID YOU GET IT FROM. and where u wake up night at 2:00 clock........................................

- 5 years ago

You Don't Need To calculate whole Terms , They will Cancel out and Only Thing left is , $_{ 2 }^{ 7 }{ C }{ e }^{ i(3x) }\quad +\quad _{ 5 }^{ 7 }{ C }{ e }^{ -i(3x) }=0$.

And what do you mean by "where u wke up night at 2.00 clock" ? Sorry But I didn't understand it.

- 5 years ago

no, you posted in 200 clock night , were you woken up. just didn't mind.. could you try the questin I have posted above .... URGENT...

- 5 years ago

Actually I ddin't Sleep , Beacuese I Used to take Sleep in Day- Evening , So I Sleep in Late Night :)

- 5 years ago

Not getting for other terms, please elaborate

- 5 years ago

Note that due to the the $2\pi$ periodicity of cosine the given problem is the same as:

$\cos^7{x}+\cos^7{(x+\frac{2\pi}{3})}+\cos^7{(x-\frac{2\pi}{3})}=0$

If only $\cos^7{x}$ were an odd function around $x$ the last two terms would cancel out! Note that at $\frac{\pi}{2}$, this is the case: cosine becomes odd when shifted by $\frac{\pi}{2}$.. But it just so happens that substituting $\frac{\pi}{2}$ kills the $\cos^7{x}$ term as well! Therefore, $x = \frac{\pi}{2}$ is a solution.

- 5 years ago

wrong ! You Calculate only one Solution ! i.e pi/2 Satisfy , But it in not only solution Recheck ur Thinking (Terms will not cancel out)

- 5 years ago

Same thing happens at every increment of $\pi$ after that, so $\frac{\pi}{2}+n\pi$ for all integers $n$. I don't think that's an exhaustive set of solutions, might come back to this tomorrow.

- 5 years ago

- 5 years ago

It is not Clearly Stated ,

But As far as I can compile This question I'am getting , velocity as a function of theta as: $\displaystyle{{ v }=\sqrt { \cfrac { 2Rg }{ 4{ \mu }^{ 2 }+1 } ((2{ { \mu } }^{ 2 }+1)\cos { \theta } -{ \mu }\sin { \theta } -(2{ { \mu } }^{ 2 }+1){ e }^{ -2\mu \theta }) } }$.

and $\displaystyle{{ v }_{ max }=\sqrt { Rg(\mu \cos { \theta } -\sin { \theta } ) } }$.

Now Putting This in velocity function we should get $\theta$ and Then We calculate distance By using fact that

$S=R\theta$.

But it is Too nasty ! Am I correctly understand ur Question ?

And Seriously It is Level-3 ??

- 5 years ago

maybe he means to say that the acceleration is so slow that almost all of friction is spent in providing the centripetal force in which it case it surely becomes level 3 ,

But either way, can you please show me how you solved the differential equation, to find velocity as function of angle

i believe that you have also done it in the same way by equating the resultant of friction to the net centripetal and tangential acceleration, but after that how you proceeded to solve it,

- 5 years ago

I'am Really Did not understand the Language , But Still I'am trying in this way ......

$\displaystyle{mg\cos { \theta } -N=\cfrac { { mv }^{ 2 } }{ R } \quad (1)\\ ds=Rd\theta \quad (2)\\ \mu N-mg\sin { \theta } =mv\cfrac { dv }{ ds } \\ \\ \quad \quad \quad 2\mu N-2mg\sin { \theta } =m\cfrac { d({ v }^{ 2 }) }{ Rd\theta } \quad \quad \quad (3)}$.

from here we get V=f($\theta$) and at V = max , accleration=0

Am I correctly understand This question If not , then What does This question Really Means ?

- 5 years ago

Bro i believe you have assumed that he is travelling in a vertical circle, maybe the question means horizontal circle

- 5 years ago

ohh :O , fish !

Lol I have Done unnecessary Calculations for vertical circle ! !$\ddot\smile$

But oK , if Now we create new question in which we consider vertical Circle , Then Is it correct ?

- 5 years ago

I Think so, It seems correct :)

- 5 years ago

sorry ,my bad.i forgot to tell you.I have edited it now

- 5 years ago

it is horizontal

- 5 years ago

i dont know how to do this i found it on a book(physics today) but i think it is like what saketh said

- 5 years ago