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Question Number 69458 by mhmd last updated on 23/Sep/19 $${find}\:{the}\:{value}\:{sin}\left(−\mathrm{13}\pi/\mathrm{6}\right)\:,\:{cos}\left(\mathrm{49}\pi/\mathrm{4}\right) \\ $$$$ \\ $$ Commented by kaivan.ahmadi last updated on 23/Sep/19 $${sin}\left(−\mathrm{2}\pi−\frac{\pi}{\mathrm{6}}\right)={sin}\left(−\frac{\pi}{\mathrm{6}}\right)=−{sin}\left(\frac{\pi}{\mathrm{6}}\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${cos}\left(\frac{\mathrm{49}\pi}{\mathrm{4}}\right)={cos}\left(\mathrm{12}\pi+\frac{\pi}{\mathrm{4}}\right)={cos}\left(\frac{\pi}{\mathrm{4}}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\…
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Question Number 69456 by TawaTawa last updated on 23/Sep/19 Commented by TawaTawa last updated on 23/Sep/19 $$\mathrm{Help}\:\mathrm{with}\:\mathrm{number}\:\mathrm{5}\:\mathrm{too}\:\mathrm{sir}.\: \\ $$ Answered by mr W last updated…
Question Number 69457 by mhmd last updated on 23/Sep/19 $${find}\:{value}\:{log}\mathrm{40}/\mathrm{9}\:\:+\mathrm{4}{log}\mathrm{5}\:\:+\mathrm{2}{log}\mathrm{6}\:\:? \\ $$ Answered by MJS last updated on 23/Sep/19 $$\mathrm{log}\:\frac{\mathrm{40}}{\mathrm{9}}\:+\mathrm{4log}\:\mathrm{5}\:+\mathrm{2log}\:\mathrm{6}\:= \\ $$$$=\mathrm{log}\:\mathrm{40}\:−\mathrm{log}\:\mathrm{9}\:+\mathrm{4log}\:\mathrm{5}\:+\mathrm{2log}\:\mathrm{2}\:+\mathrm{2log}\:\mathrm{3}= \\ $$$$=\mathrm{3log}\:\mathrm{2}\:+\mathrm{log}\:\mathrm{5}\:−\mathrm{2log}\:\mathrm{3}\:+\mathrm{4log}\:\mathrm{5}\:+\mathrm{2log}\:\mathrm{2}\:+\mathrm{2log}\:\mathrm{3}= \\…
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Question Number 134984 by Dwaipayan Shikari last updated on 09/Mar/21 $$\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{2}}+\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} }+\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} }{\mathrm{2}^{\mathrm{3}} }+\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} }{\mathrm{2}^{\mathrm{4}} }+… \\ $$$$ \\ $$Find in a closed…
Question Number 134987 by 0731619177 last updated on 09/Mar/21 Answered by Olaf last updated on 09/Mar/21 $$\mathrm{I}_{\mathrm{2}} \left({k}\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}\mathrm{sin}{x}\mathrm{cos}{x}}{\:\sqrt{\mathrm{1}−{k}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} {x}}}\:{dx} \\ $$$$\mathrm{I}_{\mathrm{2}} \left({k}\right)\:=\:−\frac{\mathrm{1}}{{k}^{\mathrm{2}}…
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