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 $\displaystyle{\cfrac { 2\pi }{ 3 } \& \cfrac { 4\pi }{ 3 } }$. ?

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## Comments

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TopNewest$y = cosx + isinx$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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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 !$\ddot\smile$

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why Complex Numbers Tag ? :O

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See megh solution , you will find it interesting :)

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

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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)

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

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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 $w^2$.

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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........................................

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$_{ 2 }^{ 7 }{ C }{ e }^{ i(3x) }\quad +\quad _{ 5 }^{ 7 }{ C }{ e }^{ -i(3x) }=0$.

You Don't Need To calculate whole Terms , They will Cancel out and Only Thing left is ,And what do you mean by "where u wke up night at 2.00 clock" ? Sorry But I didn't understand it.

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Not getting for other terms, please elaborate

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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.

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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)

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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.

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solve this: Click Here

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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 ??

@Visakh Radhakrishnan

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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,

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

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 ?

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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 ?

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