Question Number 68761 by Rasheed.Sindhi last updated on 15/Sep/19 $$\mathrm{Two}\:\boldsymbol{\mathrm{arcs}}\:\mathrm{having}\:\mathrm{their}\:\mathrm{centers}\:\mathrm{on}\:\mathrm{a} \\ $$$$\boldsymbol{\mathrm{circle}}\:\mathrm{are}\:\mathrm{cutting}\:\mathrm{each}\:\mathrm{other}\:\mathrm{at}\:\mathrm{a}\: \\ $$$$\mathrm{single}\:\mathrm{point}\:\mathrm{inside}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{and}\:\mathrm{thus} \\ $$$$\:\mathrm{dividing}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{in}\:\mathrm{four}\:\mathrm{regions}. \\ $$$$ \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{arcs}\:\mathrm{cut}\:\mathrm{each}\:\mathrm{other}\:\mathrm{in}\:\boldsymbol{\mathrm{a}}:\boldsymbol{\mathrm{b}}\:\&\:\boldsymbol{\mathrm{c}}:\boldsymbol{\mathrm{d}}\: \\ $$$$\mathrm{ratios}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{between}\:\mathrm{four} \\ $$$$\mathrm{regions}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{when}\:\mathrm{the}\:\mathrm{circle} \\…
Question Number 68673 by mr W last updated on 14/Sep/19 $${find}\:\boldsymbol{\mathrm{sin}}\:\mathrm{20}°=? \\ $$ Commented by MJS last updated on 15/Sep/19 $$\Rightarrow\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{use}\:\mathrm{this}\:\mathrm{to}\:\mathrm{get}\:\mathrm{sin}\:\mathrm{1}°,\:\mathrm{because} \\ $$$$\mathrm{sin}\:\mathrm{3}°\:=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{1}−\sqrt{\mathrm{3}}\right)\sqrt{\mathrm{5}+\sqrt{\mathrm{5}}}−\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{1}−\sqrt{\mathrm{5}}\right)\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\sqrt{\mathrm{2}} \\ $$$$\mathrm{and}…
Question Number 134165 by liberty last updated on 28/Feb/21 $$\mathrm{Solve}\:\sqrt{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}\:+\:\sqrt{\mathrm{1}−\mathrm{cos}\:\mathrm{x}}\:=\:\mathrm{2sin}\:\mathrm{x} \\ $$ Answered by EDWIN88 last updated on 28/Feb/21 $$\mathrm{square}\:\mathrm{both}\:\mathrm{sides} \\ $$$$\Leftrightarrow\:\mathrm{2}+\mathrm{2}\sqrt{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}\:=\:\mathrm{4sin}\:^{\mathrm{2}} \mathrm{x} \\…
Question Number 134145 by EDWIN88 last updated on 28/Feb/21 Commented by Ñï= last updated on 28/Feb/21 $${Any}\:{time}\:{any}\:\mathrm{2}\:{hands}\:{no}\:{coincide}.{I}\:{remember}. \\ $$ Commented by mr W last updated…
Question Number 68495 by Maclaurin Stickker last updated on 12/Sep/19 $$\mathrm{Whats}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{sin}\:\left(\mathrm{6}°\right)? \\ $$ Commented by Pk1167156@gmail.com last updated on 12/Sep/19 $$\mathrm{soln} \\ $$$$\mathrm{sin6}° \\ $$$$=\mathrm{sin}\left(\mathrm{36}−\mathrm{30}\right)…
Question Number 68482 by Maclaurin Stickker last updated on 11/Sep/19 $$\mathrm{sin}\:\mathrm{1}^{°} \:=\:? \\ $$ Commented by mr W last updated on 11/Sep/19 $$\mathrm{sin}\:\mathrm{1}^{°} \:=\:\mathrm{sin}\:\frac{\pi}{\mathrm{180}}\:\approx\frac{\pi}{\mathrm{180}}=\mathrm{0}.\mathrm{0174533} \\…
Question Number 68446 by 9102176137086 last updated on 10/Sep/19 $$\mathrm{cos}\:\left({x}−\mathrm{60}\right)+\mathrm{cos}\:\left({x}−\mathrm{30}\right)=\mathrm{sin}\:{x} \\ $$$${prove} \\ $$ Commented by mr W last updated on 10/Sep/19 $${not}\:{true}! \\ $$$${for}\:{x}=\mathrm{30}°:…
Question Number 133976 by liberty last updated on 26/Feb/21 $$\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{5}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{7}}\right)=? \\ $$ Answered by bemath last updated on 26/Feb/21 $$\mathrm{let}\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{5}}\right)=\vartheta\:\Rightarrow\begin{cases}{\mathrm{sin}\:\vartheta=\frac{\mathrm{3}}{\mathrm{5}}}\\{\mathrm{tan}\:\vartheta=\frac{\mathrm{3}}{\mathrm{4}}}\end{cases} \\ $$$$\Rightarrow\:\mathrm{sin}^{−\mathrm{1}}…
Question Number 133950 by liberty last updated on 25/Feb/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{9}\pi}{\mathrm{10}}\right).\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{7}\pi}{\mathrm{10}}\right). \\ $$ Answered by bobhans last updated on 25/Feb/21 Terms of…
Question Number 133929 by bemath last updated on 25/Feb/21 $$\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{76}°+\mathrm{cos}\:^{\mathrm{2}} \mathrm{16}°−\mathrm{cos}\:\mathrm{76}°.\mathrm{cos}\:\mathrm{16}°=? \\ $$ Answered by liberty last updated on 25/Feb/21 $$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}{ab}…