Question Number 84136 by mathocean1 last updated on 09/Mar/20 $${show}\:{that}: \\ $$$${tan}\mathrm{3}{x}=\frac{\mathrm{3}+{tan}^{\mathrm{2}} {x}}{\mathrm{1}−\mathrm{3}{tan}^{\mathrm{2}} {x}}×{tanx} \\ $$ Answered by Rio Michael last updated on 09/Mar/20 $$\mathrm{tan3}{x}\:=\:\mathrm{tan}\left(\mathrm{2}{x}\:+\:{x}\right)\:=\:\frac{\mathrm{tan2}{x}\:+\:\mathrm{tan}{x}}{\mathrm{1}−\mathrm{tan2}{x}\mathrm{tan}{x}}…
Question Number 18588 by Joel577 last updated on 25/Jul/17 $$\mathrm{sin}\:{x}\:=\:\frac{\mathrm{2}{a}\:+\:\mathrm{3}}{{a}\:+\:\mathrm{1}} \\ $$$$\mathrm{How}\:\mathrm{many}\:{a}\:\mathrm{that}\:\mathrm{can}\:\mathrm{satisfy}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{above}? \\ $$ Answered by ajfour last updated on 25/Jul/17 $$−\mathrm{1}\leqslant\frac{\mathrm{2a}+\mathrm{3}}{\mathrm{a}+\mathrm{1}}\leqslant\mathrm{1}\:\:\:;\:\:\mathrm{a}\neq−\mathrm{1} \\…
Question Number 18584 by 433 last updated on 25/Jul/17 $${Prove} \\ $$$$\mathrm{sin}\:\left(\frac{\mathrm{3}\pi}{\mathrm{10}}\right)=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$ Answered by Tinkutara last updated on 25/Jul/17 $$\mathrm{sin}\:\mathrm{54}°\:=\:\mathrm{cos}\:\mathrm{36}°\:=\:\mathrm{1}\:−\:\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{18}° \\ $$$$=\:\mathrm{1}\:−\:\frac{\mathrm{3}\:−\:\sqrt{\mathrm{5}}}{\mathrm{4}}\:=\:\frac{\sqrt{\mathrm{5}}\:+\:\mathrm{1}}{\mathrm{4}}\:\left[\because\:\mathrm{sin}\:\mathrm{18}°\:=\:\frac{\sqrt{\mathrm{5}}\:−\:\mathrm{1}}{\mathrm{4}}\right]…
Question Number 84117 by Power last updated on 09/Mar/20 Answered by TANMAY PANACEA last updated on 09/Mar/20 $${sin}\mathrm{100}+{sin}\mathrm{20}+{sin}\mathrm{50} \\ $$$$\mathrm{2}{sin}\mathrm{60}.{cos}\mathrm{40}+{cos}\mathrm{40} \\ $$$${cos}\mathrm{40}\left(\mathrm{1}+\sqrt{\mathrm{3}}\:\right) \\ $$$$\mathrm{96}×\mathrm{2}×{sin}\mathrm{40}.{cos}\mathrm{40}+\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\mathrm{30}−{cos}\mathrm{100}\right) \\…
Question Number 18546 by Tinkutara last updated on 24/Jul/17 $$\mathrm{In}\:\Delta{ABC},\:\mathrm{tan}\frac{{A}}{\mathrm{2}}\:+\:\mathrm{tan}\frac{{B}}{\mathrm{2}}\:+\:\mathrm{tan}\frac{{C}}{\mathrm{2}}\:=\:\sqrt{\mathrm{3}}, \\ $$$$\mathrm{then}\:\Delta\:\mathrm{must}\:\mathrm{be} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Equilateral} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Isosceles} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Acute}\:\mathrm{angled} \\ $$ Commented by b.e.h.i.8.3.417@gmail.com last updated…
Question Number 18545 by Tinkutara last updated on 24/Jul/17 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solutions}\:\mathrm{of} \\ $$$$\mathrm{sin3}{x}\:+\:\mathrm{cos2}{x}\:=\:\mathrm{0}\:\mathrm{in}\:\left[\mathrm{0},\:\frac{\mathrm{3}\pi}{\mathrm{2}}\right]\:\mathrm{is} \\ $$ Answered by 433 last updated on 24/Jul/17 $$ \\ $$$$ \\…
Question Number 18523 by Tinkutara last updated on 23/Jul/17 $$\mathrm{Match}\:\mathrm{the}\:\mathrm{following} \\ $$$$\boldsymbol{\mathrm{Column}}-\boldsymbol{\mathrm{I}}\:\left(\boldsymbol{\mathrm{Trigonometric}}\:\boldsymbol{\mathrm{equation}}\right) \\ $$$$\left(\mathrm{A}\right)\:\mathrm{sin}\:\mathrm{9}\theta\:=\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}\:−\:\theta\right) \\ $$$$\left(\mathrm{B}\right)\:\mathrm{sin}\:\mathrm{5}\theta\:=\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}\:+\:\mathrm{2}\theta\right) \\ $$$$\left(\mathrm{C}\right)\:\mathrm{cos}\:\mathrm{11}\theta\:=\:\mathrm{cos}\:\mathrm{3}\theta \\ $$$$\left(\mathrm{D}\right)\:\mathrm{3}\:\mathrm{tan}\:\left(\theta\:−\:\mathrm{15}°\right)\:=\:\mathrm{tan}\:\left(\theta\:+\:\mathrm{15}°\right) \\ $$$$\boldsymbol{\mathrm{Column}}-\boldsymbol{\mathrm{II}}\:\left(\boldsymbol{\mathrm{Family}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{solutions}}\right) \\ $$$$\left(\mathrm{p}\right)\:\left(\mathrm{2}{n}\:+\:\mathrm{1}\right)\frac{\pi}{\mathrm{10}},\:{n}\:\in\:{Z} \\…
Question Number 18524 by Tinkutara last updated on 23/Jul/17 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{sin}^{\mathrm{3}} {x}\:−\:\mathrm{3sin}{x}\mathrm{cos}^{\mathrm{2}} {x}\:+\:\mathrm{2cos}^{\mathrm{3}} {x}\:=\:\mathrm{0}\:\mathrm{in} \\ $$$$\left[−\frac{\pi}{\mathrm{4}},\:\frac{\pi}{\mathrm{4}}\right]\:\mathrm{is} \\ $$ Answered by Tinkutara last updated on…
Question Number 18501 by Tinkutara last updated on 22/Jul/17 $$\mathrm{In}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{ABC} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{sin}{A}.\mathrm{sin}{B}.\mathrm{sin}{C}\:=\:\frac{\Delta}{\mathrm{2}{R}^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}\right)\:\mathrm{sin}{A}.\mathrm{sin}{B}.\mathrm{sin}{C}\:=\:\frac{{r}}{\mathrm{2}{R}}\left(\mathrm{sin}{A}\:+\:\mathrm{sin}{B}\:+\:\mathrm{sin}{C}\right) \\ $$$$\left(\mathrm{3}\right)\:{a}\mathrm{cos}{A}\:+\:{b}\mathrm{cos}{B}\:+\:{c}\mathrm{cos}{C}\:=\:\frac{{abc}}{\mathrm{2}{R}^{\mathrm{2}} } \\ $$$$\left(\mathrm{4}\right)\:\mathrm{sin}{A}.\mathrm{sin}{B}.\mathrm{sin}{C}\:=\:\frac{{R}}{\mathrm{2}{r}}\left(\mathrm{sin}{A}\:+\:\mathrm{sin}{B}\:+\:\mathrm{sin}{C}\right) \\ $$ Commented by b.e.h.i.8.3.417@gmail.com…
Question Number 84019 by mathocean1 last updated on 08/Mar/20 $${Show}\:{that}: \\ $$$$\mathrm{1}\bullet\:\:\:{tan}\mathrm{3}{x}=\frac{\mathrm{3}−{tan}^{\mathrm{2}} {x}}{\mathrm{1}−\mathrm{3}{tan}^{\mathrm{2}} {x}} \\ $$$${using}\:{cos}\mathrm{3}{x}=\mathrm{4}{cos}^{\mathrm{4}} {x}−\mathrm{3}{cosx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{sin}\mathrm{3}{x}=−\mathrm{4}{sin}^{\mathrm{3}} {x}+\mathrm{3}{sinx} \\ $$$${Thanks}… \\ $$ Commented…