Question Number 176716 by cortano1 last updated on 25/Sep/22 Answered by mr W last updated on 26/Sep/22 Commented by Tawa11 last updated on 26/Sep/22 $$\mathrm{Great}\:\mathrm{sir}…
Question Number 45639 by peter frank last updated on 14/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 15/Oct/18 $$\left.\mathrm{1}\right)\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{d}}{{dx}}\left(\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}}…
Question Number 45609 by peter frank last updated on 14/Oct/18 Commented by MrW3 last updated on 14/Oct/18 $$\frac{{b}}{\mathrm{sin}\:{B}}=\frac{{a}}{\mathrm{sin}\:{A}}=\frac{{a}}{\mathrm{sin}\:\left({B}+{C}\right)} \\ $$$$\Rightarrow{b}={a}\frac{\mathrm{sin}\:{B}}{\mathrm{sin}\:\left({B}+{C}\right)} \\ $$$${Area}=\frac{\mathrm{1}}{\mathrm{2}}{ab}\:\mathrm{sin}\:{C}=\frac{{a}^{\mathrm{2}} \mathrm{sin}\:{B}\:\mathrm{sin}\:{C}}{\mathrm{2}\:\mathrm{sin}\:\left({B}+{C}\right)} \\ $$$$…
Question Number 176668 by mnjuly1970 last updated on 24/Sep/22 Answered by a.lgnaoui last updated on 25/Sep/22 $$\frac{\mathrm{2}\pi}{\mathrm{9}}=\left(\frac{\mathrm{3}\pi}{\mathrm{9}}−\frac{\pi}{\mathrm{9}}\right);\:\:\frac{\mathrm{4}\pi}{\mathrm{9}}\:=\left(\frac{\mathrm{3}\pi}{\mathrm{9}}+\frac{\pi}{\mathrm{9}}\right);\frac{\mathrm{8}\pi}{\mathrm{9}}=\left(\pi−\frac{\pi}{\mathrm{9}}\right) \\ $$$$\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{9}}=\mathrm{cos}\:\frac{\pi}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{9}}\right)\mathrm{cos}\left(\:\frac{\pi}{\mathrm{9}}\right)−\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{9}}\right)\mathrm{cos}\:\left(\frac{\pi}{\mathrm{9}}\right)=\left(\mathrm{2cos}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{9}}\right)−\mathrm{1}\right)\mathrm{cos}\:\frac{\pi}{\mathrm{9}}−\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{9}}\right)\mathrm{cos}\left(\:\frac{\pi}{\mathrm{9}}\right) \\ $$$$=\mathrm{2cos}\:^{\mathrm{3}} \left(\frac{\pi}{\mathrm{9}}\right)−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{9}}\right)−\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{9}}\right)\right)\mathrm{cos}\:\frac{\pi}{\mathrm{9}}…
Question Number 111125 by bemath last updated on 02/Sep/20 Answered by Dwaipayan Shikari last updated on 02/Sep/20 $${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