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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TopNewest\( y = cosx + isinx\)

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

\(y + \dfrac{1}{y} = 2cosx\)

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

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

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

\( = 2cos7x + 7.2cos5x + 21.2cos3x + 35.2cosx\)

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

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

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

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

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

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

\( 3x = n\pi \pm \dfrac{\pi}{2}\)

\( x = \dfrac{n\pi}{3} - \dfrac{\pi}{6}\) – Megh Choksi · 2 years ago

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– Deepanshu Gupta · 2 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 !\(\ddot\smile\)Log in to reply

do we have to use binomial expanision In the mid of the solution . plz ans my question............ and reply – Rajat Kharbanda · 2 years ago

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– Deepanshu Gupta · 2 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)Log in to reply

why Complex Numbers Tag ? :O – Karan Shekhawat · 2 years ago

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– Deepanshu Gupta · 2 years ago

See megh solution , you will find it interesting :)Log in to reply

\((e^{ix} + e^{-ix})/2 = cosx\) given equatin = 0 this implies, \(((e^{ix} + e^{-ix})/2)^{7} (1+w^2 +w) =0\) this implies for all x...... – Rajat Kharbanda · 2 years ago

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– Megh Choksi · 2 years ago

Not getting for other terms, please elaborateLog in to reply

– Deepanshu Gupta · 2 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 \(w^2\).Log in to reply

– Rajat Kharbanda · 2 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........................................Log in to reply

And what do you mean by "where u wke up night at 2.00 clock" ? Sorry But I didn't understand it. – Deepanshu Gupta · 2 years ago

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– Rajat Kharbanda · 2 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...Log in to reply

– Deepanshu Gupta · 2 years ago

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

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. – Roy Tu · 2 years ago

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– Deepanshu Gupta · 2 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)Log in to reply

– Roy Tu · 2 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.Log in to reply

solve this: Click Here – Visakh Radhakrishnan · 2 years ago

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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 – Deepanshu Gupta · 2 years ago

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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, – Mvs Saketh · 2 years ago

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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 ? – Deepanshu Gupta · 2 years ago

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– Visakh Radhakrishnan · 2 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 saidLog in to reply

– Mvs Saketh · 2 years ago

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

– Visakh Radhakrishnan · 2 years ago

it is horizontalLog in to reply

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 ? – Deepanshu Gupta · 2 years ago

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– Mvs Saketh · 2 years ago

I Think so, It seems correct :)Log in to reply

– Visakh Radhakrishnan · 2 years ago

sorry ,my bad.i forgot to tell you.I have edited it nowLog in to reply