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Question Number 207925 by lmcp1203 last updated on 30/May/24 Commented by lmcp1203 last updated on 31/May/24 $${proof}\:\alpha\:=\mathrm{20} \\ $$ Answered by Frix last updated on…
Question Number 207516 by MATHEMATICSAM last updated on 17/May/24 $$\mathrm{3}\left(\mathrm{sin}\theta\:−\:\mathrm{cos}\theta\right)^{\mathrm{4}} \:+\:\mathrm{6}\left(\mathrm{sin}\theta\:+\:\mathrm{cos}\theta\right)^{\mathrm{2}} \\ $$$$+\:\mathrm{4}\left(\mathrm{sin}^{\mathrm{6}} \theta\:+\:\mathrm{cos}^{\mathrm{6}} \theta\right)\:=\:? \\ $$ Commented by A5T last updated on 17/May/24 $${Did}\:{you}\:{edit}\:{this}\:{question}\:{to}\:{change}\:{something}?…
Question Number 207434 by Frix last updated on 22/May/24 $$\mathrm{Relating}\:\mathrm{to}\:\mathrm{question}\:\mathrm{207407} \\ $$$${x}^{\mathrm{3}} −\mathrm{12}{x}^{\mathrm{2}} +\mathrm{27}{x}−\mathrm{17}=\mathrm{0} \\ $$$$\mathrm{Let}\:{x}={t}+\mathrm{4} \\ $$$${t}^{\mathrm{3}} −\mathrm{21}{t}−\mathrm{37}=\mathrm{0} \\ $$$$\mathrm{The}\:\mathrm{Trigonometric}\:\mathrm{Solution}\:\mathrm{gives}\:\mathrm{these}: \\ $$$${x}_{\mathrm{1}} =\mathrm{4}−\mathrm{2}\sqrt{\mathrm{7}}\mathrm{cos}\:\frac{\pi+\mathrm{2sin}^{−\mathrm{1}} \:\frac{\mathrm{37}\sqrt{\mathrm{7}}}{\mathrm{98}}}{\mathrm{6}}…
Question Number 207065 by manxsol last updated on 05/May/24 $$ \\ $$$$\:\:\:{f}\left({x}\right)=\left[{cos}\mathrm{2}{x}+{cos}\mathrm{3}{x}\right]\left[{cos}\mathrm{4}{x}+{cos}\mathrm{6}{x}\right]\left[\left[{cosx}+{cos}\mathrm{5}{x}\right]\right. \\ $$$${evaluar}\:\:\:{f}\left(\frac{\mathrm{2}\pi}{\mathrm{13}}\right)\:\: \\ $$ Answered by Berbere last updated on 06/May/24 $$\left.\mathrm{4}{cos}\left({x}\right){cos}\left(\mathrm{5}{x}\right){cos}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{3}{x}\right).\left[{cos}\left(\mathrm{2}{x}\right)+{cos}\left(\:\mathrm{3}{x}\right)\right]\right]{a}\mathrm{2}\left(\right. \\…
Question Number 206970 by MATHEMATICSAM last updated on 02/May/24 $$\mathrm{If}\:\mathrm{sin}\theta\:=\:\frac{{m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}}{{m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}\:+\:\mathrm{2}{n}^{\mathrm{2}} }\:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{that}\:\mathrm{tan}\theta\:=\:\frac{{m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}}{\mathrm{2}{mn}\:+\:\mathrm{2}{n}^{\mathrm{2}} }\:. \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 206971 by MATHEMATICSAM last updated on 02/May/24 $$\mathrm{Construct}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{whose}\:\mathrm{sine}\:\mathrm{is} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}\:+\:\sqrt{\mathrm{5}}}\:. \\ $$ Answered by Rasheed.Sindhi last updated on 02/May/24 $$\mathrm{sin}\theta=\:\frac{\mathrm{3}}{\mathrm{2}\:+\:\sqrt{\mathrm{5}}} \\ $$$$\bullet\theta\:{is}\:{an}\:{angle}\:{of}\:{a}\:{right}\:{triangle} \\…
Question Number 206899 by MATHEMATICSAM last updated on 29/Apr/24 $$\mathrm{If}\:\mathrm{tan}\theta\:=\:\frac{\mathrm{2}{x}\left({x}\:+\:\mathrm{1}\right)}{\mathrm{2}{x}\:+\:\mathrm{1}}\:\mathrm{then}\:\mathrm{find}\:\mathrm{sin}\theta\:\mathrm{and} \\ $$$$\mathrm{cos}\theta. \\ $$ Answered by mathzup last updated on 29/Apr/24 $$\mathrm{1}+{tan}^{\mathrm{2}} \theta\:=\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \theta}\:\Rightarrow{cos}^{\mathrm{2}} \theta\:=\frac{\mathrm{1}}{\mathrm{1}+{tan}^{\mathrm{2}}…
Question Number 206874 by lmcp1203 last updated on 29/Apr/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 206873 by lmcp1203 last updated on 29/Apr/24 Terms of Service Privacy Policy Contact: info@tinkutara.com