Menu Close

Author: Tinku Tara

nice-calculus-n-0-cos-n-x-cos-nx-solution-1-2-n-0-cos-n-1-x-cos-x-nx-cos-x-nx-2-n-0-cos-n-1-x-cos-n-1-x-

Question Number 136570 by mnjuly1970 last updated on 23/Mar/21 $$\:\:\:\:\:\:\:\:\:…….{nice}\:\:…..\:\:\:{calculus}….. \\ $$$$\:\:\:\:\Omega=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{cos}^{{n}} \left({x}\right).{cos}\left({nx}\right)=? \\ $$$$\:\:\:{solution}:::: \\ $$$$\:\:\:\:\Omega=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{cos}^{{n}−\mathrm{1}} \left({x}\right)\left\{{cos}\left({x}−{nx}\right)+{cos}\left({x}+{nx}\right)\right. \\ $$$$\:\:\therefore\:\mathrm{2}\Omega=\underset{{n}=\mathrm{0}} {\overset{\infty}…

Why-does-cos-2-x-1-2-cos-2x-1-

Question Number 5495 by FilupSmith last updated on 16/May/16 $$\mathrm{Why}\:\mathrm{does}: \\ $$$$\mathrm{cos}^{\mathrm{2}} {x}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\left(\mathrm{2}{x}\right)+\mathrm{1}\right) \\ $$ Commented by Yozzii last updated on 16/May/16 $${Let}\:\boldsymbol{{a}}=\begin{pmatrix}{{cosA}}\\{{sinA}}\end{pmatrix}\:,\boldsymbol{{b}}=\begin{pmatrix}{{cosB}}\\{−{sinB}}\end{pmatrix}. \\ $$$${For}\:{all}\:\mathrm{0}<{A},{B}<\mathrm{2}\pi,\:{the}\:{angle}\:{between}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{b}}\:{is}\:{A}+{B}…

nice-calculus-prove-that-0-ln-1-x-2-1-x-2-2-dx-pi-2-ln-2-pi-4-

Question Number 136555 by mnjuly1970 last updated on 23/Mar/21 $$\:\:\:\:\:\:\:\:\:….{nice}\:\:\:\:……\:\:\:{calculus}…. \\ $$$$\:\:\:{prove}\:{that} \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)−\frac{\pi}{\mathrm{4}} \\ $$$$\:\:\:\:\::::::::: \\ $$ Answered by…

ln-e-ln-e-ln-e-

Question Number 71016 by mr W last updated on 10/Oct/19 $$\boldsymbol{\mathrm{ln}}\:\left(\boldsymbol{{e}}+\boldsymbol{\mathrm{ln}}\:\left(\boldsymbol{{e}}+\boldsymbol{\mathrm{ln}}\:\left(\boldsymbol{{e}}+…\right)\right)\right)=? \\ $$ Answered by mr W last updated on 11/Oct/19 $$\boldsymbol{\mathrm{ln}}\:\left(\boldsymbol{{e}}+\boldsymbol{\mathrm{ln}}\:\left(\boldsymbol{{e}}+\boldsymbol{\mathrm{ln}}\:\left(\boldsymbol{{e}}+…\right)\right)\right)={x} \\ $$$$\boldsymbol{\mathrm{ln}}\:\left(\boldsymbol{{e}}+{x}\right)={x} \\…