Question Number 171693 by cortano1 last updated on 20/Jun/22 Answered by som(math1967) last updated on 20/Jun/22 $$\:\:{tan}\mathrm{4}{x}=\frac{\mathrm{1}+{sinx}+{cosx}}{\mathrm{1}−{sinx}+{cosx}} \\ $$$$\Rightarrow\frac{{sin}\mathrm{4}{x}}{{cos}\mathrm{4}{x}}=\frac{\mathrm{1}+{sinx}+{cosx}}{\mathrm{1}−{sinx}+{cosx}} \\ $$$$\Rightarrow{sin}\mathrm{4}{x}−{sinxsin}\mathrm{4}{x}+{sin}\mathrm{4}{xcosx} \\ $$$$\:\:\:\:\:={cos}\mathrm{4}{x}+{sinxcos}\mathrm{4}{x}+{cos}\mathrm{4}{xcosx} \\ $$$$\Rightarrow{sin}\mathrm{4}{x}+{sin}\mathrm{4}{xcosx}−{sinxcos}\mathrm{4}{x}…
Question Number 106077 by bobhans last updated on 02/Aug/20 $$\mathcal{G}\mathrm{iven}\:\underset{\mathrm{n}\:=\:\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{cos}\:^{\mathrm{2n}} \left(\theta\right)\:=\:\mathrm{5}.\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{cos}\:\mathrm{2}\theta+\mathrm{cos}\:\mathrm{4}\theta+\mathrm{cos}\:\mathrm{8}\theta+\mathrm{cos}\:\mathrm{16}\theta+\mathrm{cos} \\ $$$$\mathrm{32}\theta+…= \\ $$$$\left(\mathrm{a}\right)\:\frac{\mathrm{1}}{\mathrm{5}}\:\:\:\:\:\left(\mathrm{b}\right)\:\frac{\mathrm{3}}{\mathrm{5}}\:\:\:\:\:\left(\mathrm{c}\right)\:\mathrm{2}\:\:\:\:\:\:\left(\mathrm{d}\right)\:\mathrm{1}\:\:\:\:\:\:\left(\mathrm{e}\right)\:\mathrm{3} \\ $$ Terms of Service Privacy…
Question Number 171596 by nadovic last updated on 18/Jun/22 $${Show}\:{that}\:\frac{\mathrm{cos}\:\mathrm{70}°−\mathrm{cos}\:\mathrm{20}°}{\mathrm{sin}\:\mathrm{70}°−\mathrm{sin}\:\mathrm{20}°}\:=\:−\mathrm{1} \\ $$ Commented by mr W last updated on 18/Jun/22 $$\mathrm{cos}\:\mathrm{70}°=\mathrm{cos}\:\left(\mathrm{90}°−\mathrm{20}°\right)=\mathrm{sin}\:\mathrm{20}° \\ $$$$\mathrm{cos}\:\mathrm{20}°=\mathrm{cos}\:\left(\mathrm{90}°−\mathrm{70}°\right)=\mathrm{sin}\:\mathrm{70}° \\ $$$$\frac{\mathrm{cos}\:\mathrm{70}°−\mathrm{cos}\:\mathrm{20}°}{\mathrm{sin}\:\mathrm{70}°−\mathrm{sin}\:\mathrm{20}°}\:…
Question Number 171593 by cortano1 last updated on 18/Jun/22 $$\:{The}\:{maximum}\:{value}\:{of}\:{the} \\ $$$${expression}\:\mid\sqrt{\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{2}{a}^{\mathrm{2}} }\:−\sqrt{\mathrm{2}{a}^{\mathrm{2}} −\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}}\:\mid\: \\ $$$${where}\:{a}\:{and}\:{x}\:{real}\:{numbers}\:{is}−−− \\ $$ Commented by infinityaction last updated…
Question Number 171574 by cortano1 last updated on 18/Jun/22 Answered by mr W last updated on 18/Jun/22 Commented by mr W last updated on 18/Jun/22…
Question Number 171479 by som(math1967) last updated on 16/Jun/22 $$\:\boldsymbol{{tan}}^{\mathrm{2}} \frac{\boldsymbol{\pi}}{\mathrm{7}}\:+\boldsymbol{{tan}}^{\mathrm{2}} \frac{\mathrm{3}\boldsymbol{\pi}}{\mathrm{7}}\:+\boldsymbol{{tan}}^{\mathrm{2}} \frac{\mathrm{5}\boldsymbol{\pi}}{\mathrm{7}}=? \\ $$ Commented by som(math1967) last updated on 16/Jun/22 $$\boldsymbol{{re}}\:\boldsymbol{{post}}\:\boldsymbol{{of}}\:\:\boldsymbol{{qno}}.\mathrm{171389} \\ $$…
Question Number 105916 by bobhans last updated on 02/Aug/20 $${prove}\:{that}\:\mathrm{cot}\:{x}.\:\mathrm{cot}\:\mathrm{2}{x}\:+\mathrm{cot}\:\mathrm{2}{x}.\:\mathrm{cot}\:\mathrm{3}{x}+\mathrm{2}\:= \\ $$$$\mathrm{cot}\:{x}.\left(\mathrm{cot}\:{x}−\mathrm{cot}\:\mathrm{3}{x}\right)\: \\ $$ Answered by bemath last updated on 01/Aug/20 $$\mathrm{cot}\:{x}\:\mathrm{cot}\:\mathrm{2}{x}\:+\:\mathrm{cot}\:\mathrm{2}{x}\:\mathrm{cot}\:\mathrm{3}{x}\:=\: \\ $$$$\mathrm{cot}\:\mathrm{2}{x}\:\left(\mathrm{cot}\:{x}+\mathrm{cot}\:\mathrm{3}{x}\right)= \\…
Question Number 105861 by bemath last updated on 01/Aug/20 $$\mathcal{G}{iven}\:\begin{cases}{\mathrm{sin}\:\mathrm{2}{x}−\mathrm{sin}\:\mathrm{2}{y}=−\frac{\mathrm{5}}{\mathrm{12}}}\\{\mathrm{cos}\:\left({x}+{y}\right)=\:−\frac{\mathrm{2}}{\mathrm{3}}\mathrm{sin}\:\left({x}−{y}\right)}\end{cases} \\ $$$${where}\:\mathrm{0}\:<\:{x}−{y}<\pi. \\ $$$${find}\:{the}\:{value}\:{of}\:\mathrm{cos}\:\left({x}+{y}\right)+\mathrm{2sin}\:\left({x}−{y}\right) \\ $$ Commented by PRITHWISH SEN 2 last updated on 01/Aug/20…
Question Number 171389 by cortano1 last updated on 14/Jun/22 $$\:\:\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{7}}\right)+\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)+\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{5}\pi}{\mathrm{7}}\right)=? \\ $$ Commented by Shrinava last updated on 14/Jun/22 $$\mathrm{Or}\:\mathrm{sec}^{\mathrm{2}} \:\frac{\pi}{\mathrm{7}}\:+\:\mathrm{sec}^{\mathrm{2}} \:\frac{\mathrm{3}\pi}{\mathrm{7}}\:+\:\mathrm{sec}^{\mathrm{2}}…
Question Number 40285 by MJS last updated on 18/Jul/18 $$\mathrm{solve}\:\mathrm{for}\:\mathrm{1}\:\mathrm{period} \\ $$$$\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}\:=\mathrm{sin}\:\mathrm{2}\alpha \\ $$$$\mathrm{sin}\:\frac{\beta}{\mathrm{3}}\:=\mathrm{sin}\:\mathrm{3}\beta \\ $$$$\mathrm{cos}\:\frac{\gamma}{\mathrm{2}}\:=\mathrm{cos}\:\mathrm{2}\gamma \\ $$$$\mathrm{cos}\:\frac{\delta}{\mathrm{3}}\:=\mathrm{cos}\:\mathrm{3}\delta \\ $$$$\mathrm{tan}\:\frac{\epsilon}{\mathrm{2}}\:=\mathrm{tan}\:\mathrm{2}\epsilon \\ $$$$\mathrm{tan}\:\frac{\zeta}{\mathrm{3}}\:=\mathrm{tan}\:\mathrm{3}\zeta \\ $$ Answered…