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Question Number 147809 by mathdanisur last updated on 23/Jul/21 $$\frac{\boldsymbol{{sin}}^{\mathrm{6}} \left(\boldsymbol{\alpha}\right)\:+\:\boldsymbol{{cos}}^{\mathrm{6}} \left(\boldsymbol{\alpha}\right)\:-\:\mathrm{1}}{\boldsymbol{{sin}}^{\mathrm{4}} \left(\boldsymbol{\alpha}\right)\:-\:\boldsymbol{{sin}}^{\mathrm{2}} \left(\boldsymbol{\alpha}\right)}\:=\:? \\ $$ Answered by gsk2684 last updated on 23/Jul/21 $$\left(\mathrm{sin}\:^{\mathrm{2}} \alpha\right)^{\mathrm{3}}…
Question Number 147810 by mathdanisur last updated on 23/Jul/21 $$\boldsymbol{{cos}}\left(\boldsymbol{{x}}\right)\:\centerdot\:\boldsymbol{{cos}}\left(\mathrm{3}\boldsymbol{{x}}\right)=\boldsymbol{{cos}}\left(\mathrm{5}\boldsymbol{{x}}\right)\:\centerdot\:\boldsymbol{{cos}}\left(\mathrm{7}\boldsymbol{{x}}\right) \\ $$$$\boldsymbol{{x}}\:=\:? \\ $$ Answered by gsk2684 last updated on 23/Jul/21 $$\mathrm{2}\:\mathrm{cos}\:{x}\:\mathrm{cos}\:\mathrm{3}{x}\:=\:\mathrm{2}\:\mathrm{cos}\:\mathrm{5}{x}\:\mathrm{cos}\:\mathrm{7}{x} \\ $$$$\mathrm{cos}\:\left(\mathrm{3}{x}+{x}\right)+\mathrm{cos}\:\left(\mathrm{3}{x}−{x}\right)=\mathrm{cos}\:\left(\mathrm{7}{x}+\mathrm{5}{x}\right)+\mathrm{cos}\:\left(\mathrm{7}{x}−\mathrm{5}{x}\right) \\…
Question Number 147811 by mathdanisur last updated on 23/Jul/21 Answered by Olaf_Thorendsen last updated on 23/Jul/21 $$\mathrm{X}\:=\:\mathrm{sin}\left(\frac{\mathrm{arccos}{x}}{\mathrm{4}}\right) \\ $$$$\Rightarrow\:\mathrm{X}^{\mathrm{2}} \:=\:\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{arccos}{x}}{\mathrm{4}}\right) \\ $$$$\mathrm{X}^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{cos}\left(\mathrm{2}×\frac{\mathrm{arccos}{x}}{\mathrm{4}}\right)\right) \\…
Question Number 82265 by Ali Yousafzai last updated on 19/Feb/20 $${f}\boldsymbol{{actorize}}: \\ $$$$\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)\left({x}+\mathrm{6}\right)−\mathrm{3}{x}^{\mathrm{2}} \\ $$$$ \\ $$ Answered by MJS last updated on 19/Feb/20 $$={x}^{\mathrm{4}}…
Question Number 16723 by tawa tawa last updated on 25/Jun/17 $$\mathrm{Solve}:\:\mathrm{2}^{\mathrm{x}} \:+\:\mathrm{48}\:=\:\mathrm{16x} \\ $$ Answered by mrW1 last updated on 26/Jun/17 $$\mathrm{2}^{\mathrm{x}} =\mathrm{16}\left(\mathrm{x}−\mathrm{3}\right) \\ $$$$\frac{\mathrm{2}^{\mathrm{x}}…
Question Number 16720 by Manish Raj last updated on 25/Jun/17 $${let}\:{x}={tan}\theta\:,{so}\:\theta=\mathrm{tan}^{−\mathrm{1}} {x}\:{given}\:{that}\: \\ $$$$\mathrm{tan}^{−\mathrm{1}} \sqrt{\mathrm{1}+{x}\mathrm{2}−\mathrm{1}/{x}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 147791 by Tawa11 last updated on 23/Jul/21 $$\mathrm{If}\:\:\:\:\:\:\:\mathrm{a}\:\:+\:\:\mathrm{b}\:\:+\:\:\mathrm{c}\:\:+\:\:\mathrm{d}\:\:\:=\:\:\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{a}^{\mathrm{2}} \:\:+\:\:\mathrm{b}^{\mathrm{2}} \:\:+\:\:\mathrm{c}^{\mathrm{2}} \:\:+\:\:\mathrm{d}^{\mathrm{2}} \:\:=\:\:\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{a}^{\mathrm{3}} \:\:+\:\:\mathrm{b}^{\mathrm{3}} \:\:+\:\:\mathrm{c}^{\mathrm{3}} \:\:+\:\:\mathrm{d}^{\mathrm{3}} \:\:=\:\:\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{a}^{\mathrm{4}} \:\:+\:\:\mathrm{b}^{\mathrm{4}}…
Question Number 16699 by mrW1 last updated on 25/Jun/17 $$\mathrm{Related}\:\mathrm{to}\:\mathrm{Q16675}: \\ $$$$ \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{intersection}\:\mathrm{points} \\ $$$$\mathrm{of}\:\mathrm{graph}\:\mathrm{sin}\:\mathrm{x}=\frac{\mathrm{x}}{\mathrm{10}}. \\ $$$$ \\ $$$$\mathrm{Let}'\mathrm{s}\:\mathrm{see}\:\mathrm{sin}\:\mathrm{x}\:=\:\frac{\mathrm{x}}{\mathrm{n}}\:\mathrm{with}\:\mathrm{n}>\mathrm{1}. \\ $$$$\mathrm{For}\:\mathrm{n}\leqslant\mathrm{1}\:\mathrm{there}\:\mathrm{is}\:\mathrm{one}\:\mathrm{intersection}\:\mathrm{point}. \\ $$$$ \\…
Question Number 147737 by liberty last updated on 23/Jul/21 $$\:\underset{{n}\geqslant\mathrm{1}} {\sum}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…+\frac{\mathrm{1}}{{n}}\right)^{\mathrm{2}} =? \\ $$ Answered by ArielVyny last updated on 24/Jul/21 $$\left(\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}}\right)^{\mathrm{2}} \leqslant\Sigma\frac{\mathrm{1}}{{n}^{\mathrm{2}}…