Question Number 126720 by mathocean1 last updated on 23/Dec/20 $$ \\ $$$${show}\:{that} \\ $$$${cos}\left({a}\right)+{cos}\left({b}\right)=\mathrm{2}{cos}\left(\frac{{a}+{b}}{\mathrm{2}}\right){cos}\left(\frac{{a}−{b}}{\mathrm{2}}\right) \\ $$ Answered by mathmax by abdo last updated on 23/Dec/20…
Question Number 126712 by help last updated on 23/Dec/20 Commented by liberty last updated on 24/Dec/20 $$\:\left(\bullet\right)\:{x}+\frac{\mathrm{1}}{{x}}\:=\:{w}\:\Rightarrow{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:=\:{w}^{\mathrm{2}} −\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}\:=\:{w}^{\mathrm{2}} −\mathrm{2} \\…
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Question Number 126708 by slahadjb last updated on 23/Dec/20 $$\boldsymbol{{solve}}\:\:\:\:\:\boldsymbol{{x}}+\boldsymbol{{x}}^{\sqrt{\mathrm{2}}} =\mathrm{1} \\ $$ Commented by Dwaipayan Shikari last updated on 23/Dec/20 $$\sim\mathrm{0}.\mathrm{559793} \\ $$ Commented…
Question Number 192241 by yaslm last updated on 12/May/23 Answered by Frix last updated on 12/May/23 $${Z}_{\mathrm{1}} =\frac{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{−\mathrm{i}{x}} }{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{−\mathrm{i}{x}} }=−\frac{\mathrm{1}−\mathrm{2cos}\:{x}}{\mathrm{5}−\mathrm{4cos}\:{x}}−\frac{\mathrm{2sin}\:{x}}{\mathrm{5}−\mathrm{4cos}\:{x}}\mathrm{i} \\ $$$${Z}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{{ix}} }=\frac{\mathrm{2}\left(\mathrm{2}−\mathrm{cos}\:{x}\right)}{\mathrm{5}−\mathrm{4cos}\:{x}}+\frac{\mathrm{2sin}\:{x}}{\mathrm{5}−\mathrm{4cos}\:{x}}\mathrm{i} \\…
Question Number 192239 by mathdave last updated on 12/May/23 Commented by mathdave last updated on 12/May/23 $${can}\:{someone}\:{pls}\:{help}\:{me}\:{out}\:{it}\:{very}\:{urgent} \\ $$ Answered by aleks041103 last updated on…
Question Number 192233 by gatocomcirrose last updated on 12/May/23 $$\mathrm{show}\:\mathrm{for}\:\mathrm{all}\:\mathrm{n}\in\mathrm{N}\:\mathrm{that} \\ $$$$\mathrm{3}\left(\mathrm{1}^{\mathrm{5}} +…+\mathrm{n}^{\mathrm{5}} \right)\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{1}^{\mathrm{3}} +…+\mathrm{n}^{\mathrm{3}} \\ $$ Commented by Frix last updated on 12/May/23 $${S}_{\mathrm{5}}…
Question Number 192227 by sonukgindia last updated on 12/May/23 Answered by senestro last updated on 12/May/23 $$…\mathrm{76} \\ $$ Commented by text last updated on…
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Question Number 192220 by gatocomcirrose last updated on 12/May/23 $$ \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{if}\:\mathrm{n}\in\mathbb{N},\:\mathrm{n}>\mathrm{1}\:\mathrm{and}\:\mathrm{n}\:\mathrm{is}\:\mathrm{odd}\:\mathrm{then} \\ $$$$\:\mathrm{1}^{\mathrm{n}} +…+\left(\mathrm{n}−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{n} \\ $$$$\left(\mathrm{dont}\:\mathrm{use}\:\equiv\left(\mathrm{modn}\right)\right) \\ $$ Commented by AST last updated…