Question Number 20198 by Joel577 last updated on 24/Aug/17 $$\mathrm{Find}\:\mathrm{exact}\:\mathrm{form}\:\mathrm{of} \\ $$$$\mathrm{cos}\:\left(\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$ Answered by Tinkutara last updated on 24/Aug/17 $$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)…
Question Number 20091 by Tinkutara last updated on 21/Aug/17 $$\mathrm{Prove}\:\mathrm{that}\:\underset{{n}=\mathrm{0}} {\overset{\mathrm{3}} {\sum}}\mathrm{tan}^{\mathrm{2}} \:\frac{\left(\mathrm{2}{n}\:+\:\mathrm{1}\right)\pi}{\mathrm{16}}\:=\:\mathrm{28}. \\ $$ Answered by ajfour last updated on 21/Aug/17 $$\frac{\mathrm{sin}\:^{\mathrm{2}} {A}}{\mathrm{cos}\:^{\mathrm{2}} {A}}+\frac{\mathrm{sin}\:^{\mathrm{2}}…
Question Number 85623 by jagoll last updated on 23/Mar/20 $$\mathrm{prove}\:\mathrm{that}\: \\ $$$$\mathrm{sin}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{2x}\:=\:\frac{\mathrm{1}}{\mathrm{4sin}\:\mathrm{4x}\:\mathrm{sec}\:\mathrm{x}} \\ $$ Answered by Henri Boucatchou last updated on 23/Mar/20 $${Wrong}\:{Sir}; \\ $$$${Take}\:\:{x}=\frac{\pi}{\mathrm{4}},\:\:\mathrm{0}\neq\frac{\mathrm{1}}{\mathrm{0}}…
Question Number 151132 by dw last updated on 18/Aug/21 $$ \\ $$$${Find}\:{the}\:{minimum}\:{value}\:{of}\:\:{x}\:,\:{if}\:\:\:\alpha<\mathrm{82}^{{o}} \:. \\ $$ Commented by dw last updated on 18/Aug/21 Terms of Service…
Question Number 85582 by jagoll last updated on 23/Mar/20 $$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{9}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{9}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{9}}\right)= \\ $$ Commented by som(math1967) last updated on 23/Mar/20 $$\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{9}}\right) \\ $$ Terms of Service…
Question Number 20021 by Tinkutara last updated on 20/Aug/17 $$\mathrm{A}\:\mathrm{person}\:\mathrm{observes}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{elevation} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{peak}\:\mathrm{of}\:\mathrm{a}\:\mathrm{hill}\:\mathrm{from}\:\mathrm{a}\:\mathrm{station}\:\mathrm{to}\:\mathrm{be} \\ $$$$\alpha.\:\mathrm{He}\:\mathrm{walks}\:{c}\:\mathrm{metres}\:\mathrm{along}\:\mathrm{a}\:\mathrm{slope} \\ $$$$\mathrm{inclined}\:\mathrm{at}\:\mathrm{the}\:\mathrm{angle}\:\beta\:\mathrm{and}\:\mathrm{finds}\:\mathrm{the} \\ $$$$\mathrm{angle}\:\mathrm{of}\:\mathrm{elevation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{peak}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{hill}\:\mathrm{to}\:\mathrm{be}\:\gamma.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{height}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{peak}\:\mathrm{above}\:\mathrm{the}\:\mathrm{ground}\:\mathrm{is} \\ $$$$\frac{{c}\:\mathrm{sin}\:\alpha\:\mathrm{sin}\:\left(\gamma\:−\:\beta\right)}{\left(\mathrm{sin}\:\gamma\:−\:\alpha\right)}. \\…
Question Number 151096 by john_santu last updated on 18/Aug/21 $$ \\ $$find the center of gravity with respect to point O Commented by john_santu last…
Question Number 20019 by Tinkutara last updated on 20/Aug/17 $$\mathrm{In}\:\mathrm{any}\:\mathrm{triangle}\:{ABC},\:\mathrm{prove}\:\mathrm{that} \\ $$$${a}\:\left(\mathrm{cos}\:{C}\:−\:\mathrm{cos}\:{B}\right)\:=\:\mathrm{2}\:\left({b}\:−\:{c}\right)\:\mathrm{cos}^{\mathrm{2}} \:\frac{{A}}{\mathrm{2}} \\ $$ Answered by ajfour last updated on 20/Aug/17 $${l}.{h}.{s}.={a}\left[\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}}…
Question Number 85489 by oustmuchiya@gmail.com last updated on 22/Mar/20 $${Find}\:{all}\:{angles}\:{between}\:\mathrm{0}°\:{and}\:\mathrm{360}°,\:{for}\:{which}\:\mathrm{8}{sin}\theta=\mathrm{3}{cos}^{\mathrm{2}} \theta \\ $$ Commented by Tony Lin last updated on 22/Mar/20 $$\mathrm{8}{sin}\theta=\mathrm{3}\left(\mathrm{1}−{sin}^{\mathrm{2}} \theta\right) \\ $$$$\mathrm{3}{sin}^{\mathrm{2}}…
Question Number 85484 by oustmuchiya@gmail.com last updated on 22/Mar/20 $${show}\:{that}\:\frac{\mathrm{1}}{{sec}\theta+\mathrm{1}}+\frac{\mathrm{1}}{{sec}\theta−\mathrm{1}}\equiv\mathrm{2}{cosec}\theta{cot}\theta \\ $$ Answered by som(math1967) last updated on 18/Apr/20 $$\frac{\mathrm{sec}\:\theta−\mathrm{1}+{sec}\theta+\mathrm{1}}{\left({sec}\theta+\mathrm{1}\right)\left({sec}\theta−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{2}{sec}\theta}{{sec}^{\mathrm{2}} \theta−\mathrm{1}}=\frac{\mathrm{2}{sec}\theta}{{tan}^{\mathrm{2}} \theta}=\frac{\mathrm{2}{sec}\theta{cos}^{\mathrm{2}} \theta}{{sin}^{\mathrm{2}}…