Question Number 18723 by mondodotto@gmail.com last updated on 28/Jul/17 Answered by Tinkutara last updated on 29/Jul/17 $$\mathrm{2sin2}\theta\:+\:\mathrm{15cos2}\theta\:=\:\mathrm{10} \\ $$$$\frac{\mathrm{2sin2}\theta}{\:\sqrt{\mathrm{229}}}\:+\:\frac{\mathrm{15cos2}\theta}{\:\sqrt{\mathrm{229}}}\:=\:\frac{\mathrm{10}}{\:\sqrt{\mathrm{229}}} \\ $$$$\mathrm{cos}\left(\mathrm{2}\theta\:−\:\alpha\right)\:=\:\frac{\mathrm{10}}{\:\sqrt{\mathrm{229}}}\:,\:\mathrm{where}\:\alpha\:=\:\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{15}}{\:\sqrt{\mathrm{229}}}\right) \\ $$$$\theta\:=\:{n}\pi\:\pm\:\frac{\mathrm{5}}{\:\sqrt{\mathrm{229}}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{15}}{\:\sqrt{\mathrm{229}}}\right)…
Question Number 18721 by Tinkutara last updated on 28/Jul/17 $$\mathrm{Let}\:{ABC}\:\mathrm{and}\:{ABC}'\:\mathrm{be}\:\mathrm{two}\:\mathrm{non}- \\ $$$$\mathrm{congruent}\:\mathrm{triangles}\:\mathrm{with}\:\mathrm{sides}\:{AB}\:=\:\mathrm{4}, \\ $$$${AC}\:=\:{AC}'\:=\:\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{and}\:\mathrm{angle}\:{B}\:=\:\mathrm{30}°. \\ $$$$\mathrm{The}\:\mathrm{absolute}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{difference} \\ $$$$\mathrm{between}\:\mathrm{the}\:\mathrm{areas}\:\mathrm{of}\:\mathrm{these}\:\mathrm{triangles}\:\mathrm{is} \\ $$ Answered by ajfour last updated…
Question Number 18719 by Tinkutara last updated on 28/Jul/17 $$\mathrm{If}\:\mathrm{cos}^{\mathrm{2}} \:{x}_{\mathrm{1}} \:+\:\mathrm{cos}^{\mathrm{2}} \:{x}_{\mathrm{2}} \:+\:\mathrm{cos}^{\mathrm{2}} \:{x}_{\mathrm{3}} \:+\:\mathrm{cos}^{\mathrm{2}} \:{x}_{\mathrm{4}} \\ $$$$+\:\mathrm{cos}^{\mathrm{2}} \:{x}_{\mathrm{5}} \:=\:\mathrm{5},\:\mathrm{then}\:\mathrm{sin}\:{x}_{\mathrm{1}} \:+\:\mathrm{2sin}\:{x}_{\mathrm{2}} \:+ \\ $$$$\mathrm{3sin}\:{x}_{\mathrm{3}}…
Question Number 84245 by pew_247 last updated on 10/Mar/20 $$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 18695 by Arnab Maiti last updated on 27/Jul/17 $$\mathrm{if}\:\:\frac{\mathrm{n}\:\mathrm{tan}\theta}{\mathrm{cos}^{\mathrm{2}} \left(\alpha−\theta\right)}=\frac{\mathrm{m}\:\mathrm{tan}\left(\alpha−\theta\right)}{\mathrm{cos}^{\mathrm{2}} \theta} \\ $$$$\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{2}\theta=\alpha−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{n}−\mathrm{m}}{\mathrm{n}+\mathrm{m}}\mathrm{tan}\alpha\right) \\ $$ Answered by Tinkutara last updated…
Question Number 18693 by Arnab Maiti last updated on 27/Jul/17 $$\mathrm{prove}\:\mathrm{that}\:\mathrm{2tan}^{−\mathrm{1}} \left[\mathrm{tan}\frac{\alpha}{\mathrm{2}}\mathrm{tan}\left(\frac{\pi}{\mathrm{4}}−\frac{\beta}{\mathrm{2}}\right)\right] \\ $$$$=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{sin}\alpha\:\mathrm{cos}\beta}{\mathrm{cos}\alpha\:+\mathrm{sin}\beta}\right) \\ $$ Answered by 951172235v last updated on 01/Feb/19 $$\mathrm{tan}\:\frac{\Theta}{\mathrm{2}\:}\:=\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{tan}\:\left(\frac{\overset{−}…
Question Number 84136 by mathocean1 last updated on 09/Mar/20 $${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) \\…