Question Number 219279 by Mingma last updated on 22/Apr/25 Answered by som(math1967) last updated on 22/Apr/25 $$\:\frac{\mathrm{2}{sin}\frac{\beta+\alpha}{\mathrm{2}}{cos}\frac{\alpha−\beta}{\mathrm{2}}}{\mathrm{2sin}\:\frac{\alpha+\beta}{\mathrm{2}}{sin}\frac{\alpha−\beta}{\mathrm{2}}}=\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\:{cot}\frac{\alpha−\beta}{\mathrm{2}}={cot}\frac{\pi}{\mathrm{6}} \\ $$$$\Rightarrow\:\:\alpha−\beta=\frac{\pi}{\mathrm{3}} \\ $$$$\:\mathrm{sin}\:\mathrm{3}\alpha+\mathrm{sin}\:\beta \\ $$$$=\mathrm{2sin}\:\frac{\mathrm{3}\left(\alpha+\beta\right)}{\mathrm{2}}×\mathrm{cos}\:\frac{\mathrm{3}\left(\alpha−\beta\right)}{\mathrm{2}}…
Question Number 219304 by ea last updated on 22/Apr/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 219305 by Nicholas666 last updated on 22/Apr/25 $$ \\ $$$$\:\:\:\:\:\:{Prove}; \\ $$$$\:\:\:{I}_{\mathrm{0}} \left({x}\right)\:=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\:\pi} \:{e}^{\:{x}\:{cox}\left(\theta\right)} \:{d}\theta\:; \\ $$$$\:\:\:{x}^{\mathrm{2}} {I}_{\mathrm{0}} ^{''} \left({x}\right)\:+\:{xI}'_{\mathrm{0}} \left({x}\right)\:−\:{x}^{\mathrm{2}} {I}_{\mathrm{0}}…
Question Number 219301 by SdC355 last updated on 22/Apr/25 $${f}\left({s}\right)=\frac{\mathrm{1}}{\mathrm{2}\pi}\:\int_{−\infty} ^{+\infty} \:\:{e}^{−\boldsymbol{{i}}\omega\left({s}−\alpha\right)} \mathrm{d}\omega\: \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{2}\pi}\left[{e}^{−{st}} \:\int_{\:−\infty} ^{\:+\infty} \:\:{e}^{−\boldsymbol{{i}}\omega\left({s}−\alpha\right)} \mathrm{d}\omega\right]\mathrm{d}{s}=….? \\ $$ Terms of…
Question Number 219262 by amresh last updated on 21/Apr/25 $$\overset{} {\mathrm{E}lectric}\:\mathrm{field}\:\mathrm{strenth}\:\mathrm{at}\:\mathrm{any}\:\mathrm{point}\:\mathrm{in}\:\mathrm{the}\:\mathrm{space} \\ $$$$\mathrm{is}\:\mathrm{defined}\:\mathrm{as}\:\mathrm{the}\:\mathrm{force}\:\mathrm{per}\:\mathrm{unit}\:\mathrm{charge}\:\mathrm{at}\:\mathrm{that}\:\mathrm{point}. \\ $$$$\:\mathrm{It}\:\mathrm{is}\:\mathrm{a}\:\mathrm{vector}\:\mathrm{quantity}\:\mathrm{whose}\:\mathrm{magnitude}\:\mathrm{is} \\ $$$$\mathrm{given}\:\mathrm{by}\:\mathrm{Coulomb}^{\mathrm{s}\:\:} \:\mathrm{law}\:\mathrm{and}\:\mathrm{diection}\:\mathrm{is}\:\mathrm{in}\: \\ $$$$\mathrm{straight}\:\mathrm{line}\:\mathrm{loining}\:\mathrm{the}\:\mathrm{at}\:\mathrm{that}\:\mathrm{point}. \\ $$$$\mathrm{mathemstically} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\…
Question Number 219243 by mnjuly1970 last updated on 21/Apr/25 Commented by mr W last updated on 23/Apr/25 $${i}\:{got}\:\left({solution}\:{see}\:{below}\right) \\ $$$${y}={C}_{\mathrm{1}} \mathrm{sin}\:\left(\mathrm{tan}^{−\mathrm{1}} {x}+{C}_{\mathrm{2}} \right) \\ $$…
Question Number 219268 by MrGaster last updated on 21/Apr/25 $${f}\left({x},{y}\right)=\mathrm{ln}\int_{\mathrm{0}} ^{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } {e}^{{t}^{\mathrm{2}} } {dt},{f}\left({x}\right)\left(\mathrm{1},\mathrm{2}\right)=? \\ $$ Answered by zetamaths last updated on 21/Apr/25…
Question Number 219254 by Spillover last updated on 21/Apr/25 Answered by Hanuda354 last updated on 21/Apr/25 Answered by Spillover last updated on 21/Apr/25 Answered by…
Question Number 219255 by Spillover last updated on 21/Apr/25 Answered by A5T last updated on 21/Apr/25 $$\mathrm{WLOG},\:\mathrm{let}\:\mathrm{the}\:\mathrm{side},\mathrm{s},\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\:\mathrm{be}\:\mathrm{1} \\ $$$$\Rightarrow\mathrm{b}^{\mathrm{2}} =\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\mathrm{a}×\mathrm{b}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{4}}\overset{/\mathrm{b}^{\mathrm{2}} }…
Question Number 219267 by zetamaths last updated on 21/Apr/25 $${Une}\:{fonction}\:{P}\:{est}\:{dite}\:{quasi}\:{polynomiale}\:{s}'{il}\:{existe}\:\left({pour}\:{k}\in\mathbb{N}\:\right)\:{k}+\mathrm{1}\:{fonction}\:{periodique}\left({c}_{{i}} \right)_{{i}\in\left[\mid\mathrm{0};{k}\mid\right]} {de}\:\mathbb{Z}\:{dans}\:\mathbb{R} \\ $$$$\:{telles}\:{que}\:{P}\left({n}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{c}_{{i}} \left({n}\right){n}^{{i}} \\ $$$$\left(\mathrm{1}\right)\:{Montrez}\:{que}\:{l}'{ensemble}\:{des}\:{fonction}\:{quasi}\:{polynomiale}\:{forme}\:{un}\:\mathbb{R}−{ev}\left({real}\:{space}\:{vector}\right). \\ $$$$\left(\mathrm{2}\right){Montrez}\:{que}\:{si}\:{P},{Q}:\mathbb{Z}\rightarrow\mathbb{R}\:{sont}\:{desfonction}\:{quasi}\:{polynomiale}\:{tel}\:{que}\:{P}\left({n}\right)={Q}\left({n}\right)\:\forall{n}\in\mathbb{N}\:{alors}\:{P}={Q} \\ $$ Answered by…