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Question Number 43268 by Rauny last updated on 09/Sep/18
probably, cos nx=2^(n−1) cos^n x−n2^(n−3) cos^(n−2) +(((n−3)n)/2)2^(n−5) cos^(n−4) x… wow
$$\mathrm{probably},\:\mathrm{cos}\:{nx}=\mathrm{2}^{{n}−\mathrm{1}} \mathrm{cos}^{{n}} \:{x}−{n}\mathrm{2}^{{n}−\mathrm{3}} \mathrm{cos}^{{n}−\mathrm{2}} \: \\ $$$$+\frac{\left({n}−\mathrm{3}\right){n}}{\mathrm{2}}\mathrm{2}^{{n}−\mathrm{5}} \mathrm{cos}^{{n}−\mathrm{4}} \:{x}\ldots \\ $$$$\mathrm{wow} \\ $$
Commented by Rauny last updated on 09/Sep/18
oh thx!
$$\mathrm{oh}\:\mathrm{thx}! \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Sep/18
cosx+isinx=e^(ix) (cosx+isinx)^n =(e^(ix) )^n e^(inx) =(c+is)^n c=cosx s=sinx cosnx+isinnx=c^n +nC_1 c^(n−1) (is)+nC_2 c^(n−2) (is)^2 + nC_3 c^(n−3) (is)^3 +...+(is)^n =cos^n x+i(nC_1 cos^(n−1) x.sinx)+nC_2 cos^(n−2) x.(−sin^2 x)+ nC_3 cos^(n−3) x(−isin^3 x)+...+(isinx)^n now arranging real part and imaginary part inright hand side cosnx+isinnx=(cos^n x−nC_2 cos^(n−2) sin^2 x+nC_4 cos^(n−4) xsin^4 x+...)+ i(nC_1 cos^(n−1) xsinx−nC_3 cos^(n−3) xsin^3 x+...) so cosnx=cos^n x−nC_2 cos^(n−2) sin^2 x+nC_4 cos^(n−4) xsin^4 x−...
$${cosx}+{isinx}={e}^{{ix}} \\ $$$$\left({cosx}+{isinx}\right)^{{n}} =\left({e}^{{ix}} \right)^{{n}} \\ $$$${e}^{{inx}} =\left({c}+{is}\right)^{{n}} \:\:\:{c}={cosx}\:\:\:\:{s}={sinx} \\ $$$${cosnx}+{isinnx}={c}^{{n}} +{nC}_{\mathrm{1}} {c}^{{n}−\mathrm{1}} \left({is}\right)+{nC}_{\mathrm{2}} {c}^{{n}−\mathrm{2}} \left({is}\right)^{\mathrm{2}} + \\ $$$$\:{nC}_{\mathrm{3}} {c}^{{n}−\mathrm{3}} \left({is}\right)^{\mathrm{3}} +…+\left({is}\right)^{{n}} \\ $$$$={cos}^{{n}} {x}+{i}\left({nC}_{\mathrm{1}} {cos}^{{n}−\mathrm{1}} {x}.{sinx}\right)+{nC}_{\mathrm{2}} {cos}^{{n}−\mathrm{2}} {x}.\left(−{sin}^{\mathrm{2}} {x}\right)+ \\ $$$$\:\:{nC}_{\mathrm{3}} {cos}^{{n}−\mathrm{3}} {x}\left(−{isin}^{\mathrm{3}} {x}\right)+…+\left({isinx}\right)^{{n}} \\ $$$${now}\:{arranging}\:{real}\:{part}\:{and}\:{imaginary}\:{part}\:{inright} \\ $$$${hand}\:{side} \\ $$$${cosnx}+{isinnx}=\left({cos}^{{n}} {x}−{nC}_{\mathrm{2}} {cos}^{{n}−\mathrm{2}} {sin}^{\mathrm{2}} {x}+{nC}_{\mathrm{4}} {cos}^{{n}−\mathrm{4}} {xsin}^{\mathrm{4}} {x}+…\right)+ \\ $$$${i}\left({nC}_{\mathrm{1}} {cos}^{{n}−\mathrm{1}} {xsinx}−{nC}_{\mathrm{3}} {cos}^{{n}−\mathrm{3}} {xsin}^{\mathrm{3}} {x}+…\right) \\ $$$${so}\:{cosnx}={cos}^{{n}} {x}−{nC}_{\mathrm{2}} {cos}^{{n}−\mathrm{2}} {sin}^{\mathrm{2}} {x}+{nC}_{\mathrm{4}} {cos}^{{n}−\mathrm{4}} {xsin}^{\mathrm{4}} {x}−… \\ $$
Commented by malwaan last updated on 09/Sep/18
how?
$$\mathrm{how}? \\ $$
Commented by Rauny last updated on 09/Sep/18
If you calculate it yourself, you will know the reason.
$$\mathrm{If}\:\mathrm{you}\:\mathrm{calculate}\:\mathrm{it}\:\mathrm{yourself}, \\ $$$$\mathrm{you}\:\mathrm{will}\:\mathrm{know}\:\mathrm{the}\:\mathrm{reason}. \\ $$
Commented by malwaan last updated on 20/Oct/18
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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