Question Number 123248 by Dwaipayan Shikari last updated on 24/Nov/20
$${Find}\:{the}\:{value}\:{of}\:{sin}\left(\frac{\pi}{\mathrm{9}}\right)\:\:{in}\:{exact}\:{form} \\ $$
Commented by MJS_new last updated on 24/Nov/20
$$\mathrm{sin}\:\frac{\pi}{\mathrm{9}}\:=\mathrm{sin}\:\mathrm{20}° \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{it}'\mathrm{s}\:\mathrm{possible}.\:\mathrm{the}\:\mathrm{smallest} \\ $$$$\mathrm{angle}\:\alpha\:\mathrm{for}\:\mathrm{which}\:\mathrm{we}\:\mathrm{can}\:\mathrm{exactly}\:\mathrm{express} \\ $$$$\mathrm{sin}\:\alpha\:\mathrm{is}\:\frac{\pi}{\mathrm{60}}=\mathrm{3}°\:\Rightarrow\:\mathrm{for}\:\beta\:\mathrm{with}\:\beta\neq\frac{{n}\pi}{\mathrm{60}}\:\mathrm{or}\:\beta\neq\left(\mathrm{3}{n}\right)° \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{impossible}.\:\mathrm{I}\:\mathrm{saw}\:\mathrm{an}\:“\mathrm{expression}''\:\mathrm{for} \\ $$$$\mathrm{sin}\:\mathrm{1}°\:\mathrm{but}\:\mathrm{it}\:\mathrm{uses}\:\mathrm{Cardano}'\mathrm{s}\:\mathrm{Solution}\:\mathrm{for} \\ $$$${x}^{\mathrm{3}} +{px}+{q}=\mathrm{0}\:\mathrm{while}\:\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}<\mathrm{0}\:\mathrm{but}\:\mathrm{in}\:\mathrm{this}\:\mathrm{case} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{defined}.\:\mathrm{you}\:\mathrm{can}\:\mathrm{write}\:\mathrm{it}\:\mathrm{down}\:\mathrm{but}\:\mathrm{it} \\ $$$$\mathrm{makes}\:\mathrm{no}\:\mathrm{sense}\:\mathrm{at}\:\mathrm{all}. \\ $$
Commented by Dwaipayan Shikari last updated on 24/Nov/20
$${sin}\left(\frac{\pi}{\mathrm{9}}\right)={sin}\theta\:\: \\ $$$$\mathrm{9}\theta=\pi\:\Rightarrow{sin}\mathrm{6}\theta\:={sin}\mathrm{3}\theta\:\Rightarrow\:\mathrm{2}{cos}\mathrm{3}\theta\:=\mathrm{1}\:\:\:\:\:\:\:\:\left({sin}\theta=\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\right. \\ $$$$\mathrm{4}{t}^{\mathrm{3}} −\mathrm{3}{t}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:{t}^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{4}}{t}−\frac{\mathrm{1}}{\mathrm{8}}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{27}}\left(\frac{−\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{4}}\left(−\frac{\mathrm{1}}{\mathrm{8}}\right)^{\mathrm{2}} \:=−\frac{\mathrm{1}}{\mathrm{64}}+\frac{\mathrm{1}}{\mathrm{4}.\mathrm{64}}<\mathrm{0}\:\:\left({Invalid}\right) \\ $$$${But}\:{any}\:{other}\:{way}\:{sir}? \\ $$$$ \\ $$
Commented by MJS_new last updated on 24/Nov/20
$$\mathrm{no}\:\mathrm{other}\:\mathrm{way} \\ $$
Answered by zarminaawan last updated on 24/Nov/20
$${sin}\mathrm{20}=\mathrm{0}.\mathrm{342} \\ $$
Commented by MJS_new last updated on 24/Nov/20
$$\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:“…{in}\:{exact}\:{form}'' \\ $$
Answered by TANMAY PANACEA last updated on 24/Nov/20
$${y}={sinx} \\ $$$${y}+\bigtriangleup{y}={sin}\left({x}+\bigtriangleup{x}\right) \\ $$$${here}\:\bigtriangleup{x}=\mathrm{2}^{{o}} =\left(\frac{\pi}{\mathrm{180}}×\mathrm{2}\right)=\frac{\pi}{\mathrm{90}} \\ $$$${x}=\mathrm{18}^{{o}} \\ $$$$\frac{\bigtriangleup{y}}{\bigtriangleup{x}}\approx\frac{{dy}}{{dx}} \\ $$$$\bigtriangleup{y}=\frac{{dy}}{{dx}}\bigtriangleup{x} \\ $$$$\bigtriangleup{y}={cosx}×\left(\frac{\pi}{\mathrm{90}}\right) \\ $$$${sin}\left(\mathrm{20}^{{o}} \right)={sin}\mathrm{18}^{{o}} +\frac{\pi}{\mathrm{90}}×{cos}\mathrm{18}^{{o}} \\ $$$$=\frac{\sqrt{\mathrm{5}}\:−\mathrm{1}}{\mathrm{4}}+\left(\frac{\pi}{\mathrm{90}}\right)×\sqrt{\frac{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}}\: \\ $$$$=\mathrm{0}.\mathrm{30901699437}+\frac{\pi}{\mathrm{90}}×\mathrm{0}.\mathrm{9510565163}\:\left({using}\:{calculator}\right) \\ $$$$=\mathrm{0}.\mathrm{30901699437}+\mathrm{0}.\mathrm{03321149739} \\ $$$$=\mathrm{0}.\mathrm{34222849176} \\ $$
Commented by MJS_new last updated on 24/Nov/20
$$\mathrm{interesting}! \\ $$$$\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{also}\:\mathrm{no}\:\mathrm{exact}\:\mathrm{solution}…\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{possible} \\ $$
Commented by TANMAY PANACEA last updated on 24/Nov/20
$${thank}\:{you}\:{sir}..{yes}\:{you}\:{correct} \\ $$
Commented by Dwaipayan Shikari last updated on 24/Nov/20
$${Thanking}\:{for}\:{interaction}\:{sirs}\:! \\ $$