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Question-102635

Question Number 102635 by ketto255 last updated on 10/Jul/20 Answered by bobhans last updated on 10/Jul/20 $$\mathrm{sin}\:\left(\mathrm{90}^{{o}} −\theta\right)\:=\:\mathrm{cos}\:\theta\:=\pm\:\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \theta}\:= \\ $$$$\pm\sqrt{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{25}}}\:=\:\pm\:\frac{\sqrt{\mathrm{23}}}{\mathrm{5}} \\ $$ Answered by…

Question-37089

Question Number 37089 by behi83417@gmail.com last updated on 08/Jun/18 Commented by math khazana by abdo last updated on 09/Jun/18 $$\left.\mathrm{1}\right)\:{f}\:{id}\:{defined}\:{on}\:\left[\mathrm{0},+\infty\left[\right.\right. \\ $$$${f}^{'} \left({x}\right)=\:\frac{\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}+{x}}}\left(\mathrm{1}+\sqrt{{x}}\right)\:−\sqrt{\mathrm{1}+{x}}\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}}{\left(\mathrm{1}+\sqrt{{x}}\right)^{\mathrm{2}} } \\…

show-that-cos-cos2-cos-n-cos-1-2-n-1-sin-1-2-n-sin-1-2-n-Show-that-sin-sin-2-sin-n-sin-1-2-n-1-sin-1-2-n-sin-1-2-n-where-R-and-2pik-k

Question Number 102627 by Rio Michael last updated on 10/Jul/20 $$\mathrm{show}\:\mathrm{that}:\:\mathrm{cos}\theta\:+\:\mathrm{cos2}\theta\:+\:….\mathrm{cos}\:{n}\theta=\:\frac{\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}\left({n}\:+\mathrm{1}\right)\theta\:\mathrm{sin}\frac{\mathrm{1}}{\mathrm{2}}{n}\theta}{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{n}\theta} \\ $$$$\mathrm{Show}\:\mathrm{that}:\:\mathrm{sin}\:\theta\:+\:\mathrm{sin}\:\mathrm{2}\theta\:+\:….+\:\mathrm{sin}\:{n}\theta\:=\:\frac{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\left({n}\:+\:\mathrm{1}\right)\theta\:\mathrm{sin}\frac{\mathrm{1}}{\mathrm{2}}{n}\theta}{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{n}\theta} \\ $$$$\mathrm{where}\:\theta\:\in\:\mathbb{R}\:\mathrm{and}\:\theta\:\neq\mathrm{2}\pi{k}\:,\:{k}\:\in\mathbb{Z} \\ $$$$ \\ $$ Answered by Dwaipayan Shikari last updated…

Question-37081

Question Number 37081 by tanmay.chaudhury50@gmail.com last updated on 08/Jun/18 Commented by candre last updated on 09/Jun/18 $$\mathrm{techinaly}\:\mathrm{newton}\:\mathrm{Segond}\:\mathrm{law}\:\mathrm{is}\:\mathrm{write}\:\mathrm{as} \\ $$$${F}=\frac{\partial{p}}{\partial{t}} \\ $$$$\mathrm{but}\:\mathrm{since}\:\mathrm{most}\:\mathrm{case}\:\mathrm{mass}\:\mathrm{is}\:\mathrm{constant}\:\mathrm{you}\:\mathrm{can}\:\mathrm{simplify}\:\mathrm{to} \\ $$$${F}=\frac{\partial}{\partial{t}}\left({mv}\right)={m}\frac{\partial{v}}{\partial{t}}={ma} \\ $$…

Question-102598

Question Number 102598 by dw last updated on 10/Jul/20 Answered by mr W last updated on 10/Jul/20 $${z}=\mathrm{2}\left(\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta\right) \\ $$$$\mid{z}−{i}\mid=\sqrt{\mathrm{4cos}^{\mathrm{2}} \:\theta+\left(\mathrm{2}\:\mathrm{sin}\:\theta−\mathrm{1}\right)^{\mathrm{2}} }=\sqrt{\mathrm{5}−\mathrm{4}\:\mathrm{sin}\:\theta} \\ $$$$\mid{z}+{i}\mid=\sqrt{\mathrm{4cos}^{\mathrm{2}} \:\theta+\left(\mathrm{2}\:\mathrm{sin}\:\theta+\mathrm{1}\right)^{\mathrm{2}}…

If-tan-i-cos-isin-prove-that-n-2-4-and-1-2-log-tan-4-2-

Question Number 168118 by Mastermind last updated on 03/Apr/22 $${If}\:{tan}\left(\theta+{i}\phi\right)={cos}\alpha+{isin}\alpha,\: \\ $$$${prove}\:{that}\::\:\theta=\frac{{n}\Pi}{\mathrm{2}}+\frac{\Pi}{\mathrm{4}}\:{and}\:\phi=\frac{\mathrm{1}}{\mathrm{2}}{log}\:{tan}\left(\frac{\Pi}{\mathrm{4}}+\frac{\alpha}{\mathrm{2}}\right) \\ $$ Answered by mathsmine last updated on 05/Apr/22 $${tan}\left(\theta+{i}\emptyset\right)=\frac{{tan}\left(\theta\right)+{tan}\left({i}\emptyset\right)}{\mathrm{1}−{tan}\left(\theta\right){tan}\left({i}\emptyset\right)} \\ $$$${tan}\left({ix}\right)={ith}\left({x}\right) \\…

If-a-b-c-a-b-b-c-a-b-c-c-a-b-c-a-and-a-b-c-0-then-which-of-the-following-is-true-1-a-b-c-2-a-b-c-3-a-b-c-4-a-b-c-

Question Number 37037 by jayanta11 last updated on 08/Jun/18 $${If}\:\frac{{a}+{b}−{c}}{{a}+{b}}\:=\:\frac{{b}+{c}−{a}}{{b}+{c}}\:=\frac{{c}+{a}−{b}}{{c}+{a}}\:{and}\: \\ $$$$\mathrm{a}+\mathrm{b}+\mathrm{c}\:\neq\:\mathrm{0}\:{then}\:{which}\:{of}\:{the}\:{following} \\ $$$$\left.{i}\left.{s}\:{true}\:\mathrm{1}\right)\:{a}=−{b}={c}\:\mathrm{2}\right)−{a}=−{b}={c}\: \\ $$$$\left.\mathrm{3}\left.\right)\:\mathrm{a}=\mathrm{b}=\mathrm{c}\:\mathrm{4}\right)\:\mathrm{a}=\mathrm{b}\neq{c} \\ $$ Answered by math1967 last updated on 08/Jun/18…