Question Number 23243 by paro123 last updated on 27/Oct/17 $$\mathrm{2}^{\mathrm{n}−\mathrm{1}} \mathrm{sin}\:\mathrm{a}×\mathrm{sin}\:\mathrm{2a}×\mathrm{sin}\:\mathrm{3a}×……×\mathrm{sin}\:\left(\mathrm{n}−\mathrm{1}\right)\mathrm{a}=\mathrm{n}\:\mathrm{why}? \\ $$ Commented by mrW1 last updated on 28/Oct/17 $$\mathrm{this}\:\mathrm{statement}\:\mathrm{is}\:\mathrm{wrong}! \\ $$$$ \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{find}\:\mathrm{a}\:\mathrm{lot}\:\mathrm{of}\:\mathrm{examples}:…
Question Number 23231 by Tinkutara last updated on 27/Oct/17 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solution}\left(\mathrm{s}\right)\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$${x}^{\mathrm{3}} \:+\:{x}^{\mathrm{2}} \:+\:\mathrm{4}{x}\:+\:\mathrm{2sin}{x}\:=\:\mathrm{0}\:\mathrm{in}\:\left[\mathrm{0},\:\mathrm{2}\pi\right],\:\mathrm{is}/\mathrm{are} \\ $$ Answered by ajfour last updated on 28/Oct/17 $${One}\:. \\…
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Question Number 154240 by amin96 last updated on 15/Sep/21 $$\begin{cases}{{tg}\left({a}+{b}\right)=\mathrm{7}}\\{{tg}\left({a}−{b}\right)=\mathrm{5}}\end{cases}\:\:{tg}\left({a}\right)=?\:\:\:{easy}\:{question} \\ $$ Answered by mr W last updated on 15/Sep/21 $${a}+{b}=\mathrm{tan}^{−\mathrm{1}} \mathrm{7} \\ $$$${a}−{b}=\mathrm{tan}^{−\mathrm{1}} \mathrm{5}…
Question Number 154216 by EDWIN88 last updated on 15/Sep/21 $${Solve}\:{the}\:{equation}\:{for}\:{reals}\: \\ $$$$\:\frac{\mathrm{cos}\:^{\mathrm{2}} {x}−\frac{\mathrm{1}}{\mathrm{5}}\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} {x}−\frac{\mathrm{1}}{\mathrm{5}}\mathrm{cos}\:^{\mathrm{2}} {x}+\mathrm{1}}\:=\:\mathrm{5tan}\:^{\mathrm{2}} {x}\: \\ $$ Answered by MJS_new last updated on…
Question Number 88613 by ajfour last updated on 11/Apr/20 Commented by ajfour last updated on 11/Apr/20 $${Find}\:{a}/{b}\:. \\ $$ Commented by ajfour last updated on…
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Question Number 88567 by jagoll last updated on 11/Apr/20 $$\left(\mathrm{1}+\mathrm{cos}\:\frac{\pi}{\mathrm{11}}\right)\left(\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{11}}\right)\left(\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{11}}\right)\left(\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{11}}\right)\left(\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{9}\pi}{\mathrm{11}}\right) \\ $$ Commented by jagoll last updated on 11/Apr/20 $$\mathrm{T}=\:\left(\mathrm{2cos}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{22}}\right)\right)\left(\mathrm{2cos}^{\mathrm{2}} \:\left(\frac{\mathrm{3}\pi}{\mathrm{22}}\right)\right)\left(\mathrm{2cos}\:^{\mathrm{2}} \left(\frac{\mathrm{5}\pi}{\mathrm{22}}\right)\right) \\ $$$$\left(\mathrm{2cos}\:^{\mathrm{2}}…
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Question Number 23009 by ajfour last updated on 25/Oct/17 Commented by ajfour last updated on 25/Oct/17 $${For}\:\:\:{Q}.\mathrm{22929} \\ $$ Commented by selestian last updated on…