Question Number 111125 by bemath last updated on 02/Sep/20 Answered by Dwaipayan Shikari last updated on 02/Sep/20 $${cos}\mathrm{55}°{cos}\mathrm{65}°{cos}\mathrm{175}° \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{cos}\mathrm{55}°{cos}\mathrm{5}°\right){cos}\mathrm{65}° \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\mathrm{60}°+{cos}\mathrm{50}°\right){sin}\mathrm{25}° \\ $$$$−\frac{\mathrm{1}}{\mathrm{4}}{sin}\mathrm{25}°−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}{cos}\mathrm{50}°{sin}\mathrm{25}°\right) \\…
Question Number 111114 by bobhans last updated on 02/Sep/20 $$\mathrm{4}\:\mathrm{sin}\:\mathrm{36}°\:\mathrm{cos}\:\mathrm{72}°\:\mathrm{sin}\:\mathrm{108}°\:?\: \\ $$ Answered by bemath last updated on 02/Sep/20 $$\:\:\sqrt{\mathrm{bemath}} \\ $$$$\:\mathrm{we}\:\mathrm{want}\:\mathrm{to}\:\mathrm{compute}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\mathrm{4}\:\mathrm{sin}\:\mathrm{36}°\:\mathrm{cos}\:\mathrm{72}°\:\mathrm{sin}\:\mathrm{108}°\:. \\…
Question Number 111079 by shahria14 last updated on 02/Sep/20 Answered by mr W last updated on 02/Sep/20 $${t}=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{5}}{\mathrm{13}} \\ $$$$\Rightarrow\mathrm{cos}\:{t}=\frac{\mathrm{5}}{\mathrm{13}} \\ $$$$\Rightarrow\mathrm{sin}\:{t}=\frac{\mathrm{12}}{\mathrm{13}} \\ $$$$\mathrm{tan}\:\frac{{t}}{\mathrm{2}}=\frac{\mathrm{sin}\:{t}}{\mathrm{1}+\mathrm{cos}\:{t}}=\frac{\frac{\mathrm{12}}{\mathrm{13}}}{\mathrm{1}+\frac{\mathrm{5}}{\mathrm{13}}}=\frac{\mathrm{2}}{\mathrm{3}}…
Question Number 176603 by a.lgnaoui last updated on 23/Sep/22 $${look}\:{the}\:{anser} \\ $$ Commented by a.lgnaoui last updated on 23/Sep/22 Commented by mr W last updated…
Question Number 176581 by cortano1 last updated on 22/Sep/22 $$\:\:\mathrm{Given}\:\begin{cases}{\mathrm{sin}\:\mathrm{a}+\mathrm{sin}\:\mathrm{b}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\\{\mathrm{cos}\:\mathrm{a}+\mathrm{cos}\:\mathrm{b}=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}}\end{cases} \\ $$$$\:\mathrm{for}\:\mathrm{a},\mathrm{b}\:\mathrm{real}\:\mathrm{numbers}.\:\mathrm{Evaluate} \\ $$$$\:\mathrm{sin}\:\left(\mathrm{a}+\mathrm{b}\right). \\ $$$$\:\left(\mathrm{A}\right)\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\:\left(\mathrm{B}\right)\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\:\:\:\:\left(\mathrm{E}\right)−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}} \\ $$$$\:\:\left(\mathrm{C}\right)\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$ Answered by som(math1967)…
Question Number 45511 by Tawa1 last updated on 13/Oct/18 $$\mathrm{In}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{ABC}\:,\:\:\angle\:\mathrm{ABC}\:=\:\mathrm{30}^{\mathrm{0}} \:,\:\:\mathrm{and}\:\:\mathrm{AC}\:=\:\mathrm{10}.\:\mathrm{A}\:\mathrm{circle}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{to} \\ $$$$\mathrm{circumscribe}\:\mathrm{the}\:\mathrm{triangle}\:.\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 111002 by bemath last updated on 01/Sep/20 $$\sqrt{\mathrm{bemath}} \\ $$$$\Rightarrow\:\mathrm{sin}\:\mathrm{14}°+\mathrm{cos}\:\mathrm{14}°\mathrm{tan}\:\mathrm{38}°−\mathrm{1}=? \\ $$ Commented by Khanacademy last updated on 01/Sep/20 $$\alpha\boldsymbol{{beMath\_uz}}\:\:\:\boldsymbol{{sizning}}\:\:\boldsymbol{{kanalingizmi}} \\ $$ Answered…
Question Number 45440 by Rio Michael last updated on 13/Oct/18 $${show}\:{that}\: \\ $$$$\frac{\mathrm{1}+\mathrm{2}{sin}\mathrm{2}\theta−{cos}\mathrm{2}\theta}{\mathrm{1}+{sin}\mathrm{2}\theta+{cos}\mathrm{2}\theta}\:\equiv\:{tan}\theta \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 13/Oct/18 $$\frac{\mathrm{1}−{cos}\mathrm{2}\theta+{sin}\mathrm{2}\theta}{\mathrm{1}+{cos}\mathrm{2}\theta+{sin}\mathrm{2}\theta} \\ $$$$\frac{\mathrm{2}{sin}^{\mathrm{2}}…
Question Number 176490 by mnjuly1970 last updated on 20/Sep/22 $$ \\ $$$$\:\:\:\:\:{solve}\:\:\:\left(\:{x},{y}\:\in\:\mathbb{R}\:\right) \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\begin{cases}{\:\mathrm{tan}\left({x}\:\right)\:+\:\mathrm{tan}\:\left({y}\:\right)=\mathrm{2}}\\{\:\mathrm{tan}\left(\mathrm{2}{x}\:\right)\:+\:\mathrm{tan}\left(\:\mathrm{2}{y}\:\right)\:=\:\mathrm{2}}\end{cases} \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−−−−−−−− \\ $$ Answered by ajfour…
Question Number 176468 by blackmamba last updated on 20/Sep/22 $$\:\:\frac{\mathrm{sin}\:\left(\mathrm{2}{x}+\mathrm{18}°\right)}{\mathrm{sin}\:\left(\mathrm{2}{x}+\mathrm{12}°\right)}\:=\sqrt{\frac{\mathrm{sin}\:\mathrm{36}°}{\mathrm{sin}\:\mathrm{48}°}}\: \\ $$$$\:\:\mathrm{tan}\:\mathrm{2}{x}\:=\:\sqrt{\mathrm{tan}\:{M}}\:.\sqrt{\mathrm{tan}\:{N}} \\ $$$$\:\:\mathrm{0}°<{M},{N}<\mathrm{90}°\:\Rightarrow{M}+{N}=?° \\ $$ Commented by Peace last updated on 20/Sep/22 $${i}\:{started}\:{Withe}\:\mathrm{48}\:,{i}\:{see}\:{no}\:{nice}\:{Form} \\…