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Category: Algebra

Question-146096

Question Number 146096 by mathdanisur last updated on 10/Jul/21 Answered by mindispower last updated on 10/Jul/21 $${tg}\left(\mathrm{2}{a}\right)={tg}\left({a}+{b}+{a}−{b}\right)=\frac{{tg}\left({a}+{b}\right)+{tg}\left({a}−{b}\right)}{\mathrm{1}−{tg}\left({a}+{b}\right){tg}\left({a}−{b}\right)}=\frac{\frac{\mathrm{3}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\mathrm{3}\Rightarrow\mathrm{2}{tg}\left(\mathrm{2}{a}\right)=\mathrm{6} \\ $$ Commented by mathdanisur last…

Simplify-sin-3-1-cos-cos-3-sin-1-

Question Number 146044 by mathdanisur last updated on 10/Jul/21 $${Simplify}: \\ $$$$\frac{{sin}^{\mathrm{3}} \alpha}{\mathrm{1}-{cos}\alpha}\:+\:\frac{{cos}^{\mathrm{3}} \alpha}{{sin}\alpha+\mathrm{1}}\:=\:? \\ $$ Answered by liberty last updated on 10/Jul/21 $$\:\Rightarrow\:\frac{\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \alpha\right)\mathrm{sin}\:\alpha}{\mathrm{1}−\mathrm{cos}\:\alpha}\:+\:\frac{\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}}…

f-x-x-sin-x-f-x-

Question Number 146046 by mathdanisur last updated on 10/Jul/21 $${f}\left({x}\right)\:=\:{x}^{\boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)} \:\Rightarrow\:{f}\:^{'} \left({x}\right)\:=\:? \\ $$ Answered by liberty last updated on 10/Jul/21 $$\:\mathrm{ln}\:{f}\left({x}\right)=\mathrm{sin}\:{x}.\:\mathrm{ln}\:\left({x}\right) \\ $$$$\Leftrightarrow\:\frac{{f}\:'\left({x}\right)}{{f}\left({x}\right)}=\:\mathrm{cos}\:{x}.\mathrm{ln}\:\left({x}\right)+\frac{\mathrm{sin}\:{x}}{{x}} \\…

Solve-for-a-b-and-c-a-b-c-1-2-i-abc-1-4-iii-ab-ac-bc-3-2-iv-

Question Number 80493 by TawaTawa last updated on 03/Feb/20 $$\mathrm{Solve}\:\mathrm{for}\:\:\mathrm{a},\:\mathrm{b}\:\mathrm{and}\:\mathrm{c} \\ $$$$\:\:\:\:\:\:\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}\:\:=\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:…..\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{abc}\:\:\:=\:\:\:−\:\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:……\:\left(\mathrm{iii}\right) \\ $$$$\:\:\:\:\:\:\mathrm{ab}\:+\:\mathrm{ac}\:+\:\mathrm{bc}\:\:\:=\:\:\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\:\:\:\:……\:\left(\mathrm{iv}\right) \\ $$ Commented by mr W last updated on…

1-4-1-12-1-24-1-2n-n-1-

Question Number 146009 by iloveisrael last updated on 10/Jul/21 $$\:\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{24}}+…+\frac{\mathrm{1}}{\mathrm{2n}\left(\mathrm{n}+\mathrm{1}\right)}=? \\ $$ Answered by puissant last updated on 10/Jul/21 $$=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{2k}\left(\mathrm{k}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}}…