Question Number 57474 by malwaan last updated on 05/Apr/19
$$\mathrm{prove}\: \\ $$$$\boldsymbol{{sin}}\mathrm{18}×\boldsymbol{{cos}}\mathrm{36}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by Abdo msup. last updated on 05/Apr/19
$$\mathrm{18}\rightarrow{x} \\ $$$$\mathrm{180}\rightarrow\pi\:\Rightarrow\mathrm{180}{x}\:=\mathrm{18}\pi\:\Rightarrow{x}\:=\frac{\pi}{\mathrm{10}}\:\Rightarrow\mathrm{36}\:=\frac{\pi}{\mathrm{5}} \\ $$$${for}\:{that}\:{let}\:{prove}\:{sin}\left(\frac{\pi}{\mathrm{10}}\right){cos}\left(\frac{\pi}{\mathrm{5}}\right)=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${we}\:{have}\:{proved}\:{that}\:{cos}\left(\frac{\pi}{\mathrm{5}}\right)=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$${sin}\left(\frac{\pi}{\mathrm{10}}\right)=\sqrt{\frac{\mathrm{1}−{cos}\left(\frac{\pi}{\mathrm{5}}\right)}{\mathrm{2}}}=\sqrt{\frac{\mathrm{1}−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}}{\mathrm{2}}} \\ $$$$=\sqrt{\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{8}}}=\frac{\sqrt{\mathrm{3}−\sqrt{\mathrm{5}}}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$${we}\:{have}\:\left\{\:{sin}\left(\frac{\pi}{\mathrm{10}}\right){cos}\left(\frac{\pi}{\mathrm{5}}\right)\right\}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{8}}\:\frac{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{16}}\:=\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{8}}\:\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{8}}\:=\frac{\mathrm{9}−\mathrm{5}}{\mathrm{64}}\:=\frac{\mathrm{4}}{\mathrm{64}}\:=\frac{\mathrm{1}}{\mathrm{16}}\:\Rightarrow \\ $$$${sin}\left(\frac{\pi}{\mathrm{10}}\right).{cos}\left(\frac{\pi}{\mathrm{5}}\right)\:=\frac{\mathrm{1}}{\mathrm{4}}\:. \\ $$
Commented by malwaan last updated on 06/Apr/19
$$\boldsymbol{{thank}}\:\boldsymbol{{you}}\:\boldsymbol{{sir}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 05/Apr/19
![a=18^o 5a=2a+3a=90^o 2a=90^o −3a sin2a=sin(90^o −3a) 2sinacosa=cos3a 2sinacosa=4cos^3 a−3cosa cosa(2sina−4cos^2 a+3)=0 cosa≠0 2sina−4(1−sin^2 a)+3=0 4sin^2 a+2sina−1=0 sin18^o >0 [ 18^o lies in first quadrant] sina=((−2+(√((2)^2 −4×4×(−1))))/(2×4)) sin18^o =((−2+2(√5))/8) sin18^o =(((√5) −1)/4) cos36^o =1−2sin^2 18^o =1−2((((√5) −1)/4))^2 =((16−2(5−2(√5) +1))/(16))=((8−(6−2(√5) ))/8) =((2(√5) +2)/8)→(((√5) +1)/4) sin18×cos36 =(((√5) −1)/4)×(((√5) +1)/4) =((5−1)/(16))=(1/4) i have derived the value of sin18^o and cos36^o then solved the problem.. for short cut method memorize the value of sin18^o and cos36^o then solve it...](https://www.tinkutara.com/question/Q57476.png)
$${a}=\mathrm{18}^{{o}} \\ $$$$\mathrm{5}{a}=\mathrm{2}{a}+\mathrm{3}{a}=\mathrm{90}^{{o}} \\ $$$$\mathrm{2}{a}=\mathrm{90}^{{o}} −\mathrm{3}{a} \\ $$$${sin}\mathrm{2}{a}={sin}\left(\mathrm{90}^{{o}} −\mathrm{3}{a}\right) \\ $$$$\mathrm{2}{sinacosa}={cos}\mathrm{3}{a} \\ $$$$\mathrm{2}{sinacosa}=\mathrm{4}{cos}^{\mathrm{3}} {a}−\mathrm{3}{cosa} \\ $$$${cosa}\left(\mathrm{2}{sina}−\mathrm{4}{cos}^{\mathrm{2}} {a}+\mathrm{3}\right)=\mathrm{0} \\ $$$${cosa}\neq\mathrm{0} \\ $$$$\mathrm{2}{sina}−\mathrm{4}\left(\mathrm{1}−{sin}^{\mathrm{2}} {a}\right)+\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{4}{sin}^{\mathrm{2}} {a}+\mathrm{2}{sina}−\mathrm{1}=\mathrm{0} \\ $$$${sin}\mathrm{18}^{{o}} \:>\mathrm{0}\:\:\left[\:\mathrm{18}^{{o}} \:{lies}\:{in}\:{first}\:{quadrant}\right] \\ $$$${sina}=\frac{−\mathrm{2}+\sqrt{\left(\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}×\mathrm{4}×\left(−\mathrm{1}\right)}}{\mathrm{2}×\mathrm{4}} \\ $$$${sin}\mathrm{18}^{{o}} =\frac{−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{8}} \\ $$$${sin}\mathrm{18}^{{o}} =\frac{\sqrt{\mathrm{5}}\:\:−\mathrm{1}}{\mathrm{4}} \\ $$$${cos}\mathrm{36}^{{o}} =\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \mathrm{18}^{{o}} \\ $$$$=\mathrm{1}−\mathrm{2}\left(\frac{\sqrt{\mathrm{5}}\:−\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{16}−\mathrm{2}\left(\mathrm{5}−\mathrm{2}\sqrt{\mathrm{5}}\:+\mathrm{1}\right)}{\mathrm{16}}=\frac{\mathrm{8}−\left(\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}\:\right)}{\mathrm{8}} \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{5}}\:+\mathrm{2}}{\mathrm{8}}\rightarrow\frac{\sqrt{\mathrm{5}}\:+\mathrm{1}}{\mathrm{4}} \\ $$$${sin}\mathrm{18}×{cos}\mathrm{36} \\ $$$$=\frac{\sqrt{\mathrm{5}}\:−\mathrm{1}}{\mathrm{4}}×\frac{\sqrt{\mathrm{5}}\:+\mathrm{1}}{\mathrm{4}} \\ $$$$=\frac{\mathrm{5}−\mathrm{1}}{\mathrm{16}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${i}\:{have}\:{derived}\:{the}\:{value}\:{of}\:{sin}\mathrm{18}^{{o}} \:{and}\:{cos}\mathrm{36}^{{o}} \\ $$$${then}\:{solved}\:{the}\:{problem}.. \\ $$$${for}\:{short}\:{cut}\:{method}\:{memorize}\:{the}\:{value} \\ $$$${of}\:{sin}\mathrm{18}^{{o}} \:{and}\:{cos}\mathrm{36}^{{o}} \:{then}\:{solve}\:{it}… \\ $$
Commented by malwaan last updated on 06/Apr/19
$$\boldsymbol{{thank}}\:\boldsymbol{{you}}\:\boldsymbol{{so}}\:\boldsymbol{{much}} \\ $$