Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 109923 by mathdave last updated on 26/Aug/20

find   sin3     in surd form

$${find}\: \\ $$$${sin}\mathrm{3}\:\:\: \\ $$$${in}\:{surd}\:{form} \\ $$

Answered by malwan last updated on 26/Aug/20

sin 3° = sin (18−15)  =sin 18 cos15 −cos18 sin15  =((((√5)−1)/4))(((√(2+(√3)))/2))−(((√(5+(√5)))/2))(((√(2−(√3)))/2))  =(1/(48))(√6)((√5)−1)(3+(√3))−     (1/(24))((√3)(3−(√3))((√(5+(√5))))

$${sin}\:\mathrm{3}°\:=\:{sin}\:\left(\mathrm{18}−\mathrm{15}\right) \\ $$$$={sin}\:\mathrm{18}\:{cos}\mathrm{15}\:−{cos}\mathrm{18}\:{sin}\mathrm{15} \\ $$$$=\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\right)\left(\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}{\mathrm{2}}\right)−\left(\frac{\sqrt{\mathrm{5}+\sqrt{\mathrm{5}}}}{\mathrm{2}}\right)\left(\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{48}}\sqrt{\mathrm{6}}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)\left(\mathrm{3}+\sqrt{\mathrm{3}}\right)− \\ $$$$\:\:\:\frac{\mathrm{1}}{\mathrm{24}}\left(\sqrt{\mathrm{3}}\left(\mathrm{3}−\sqrt{\mathrm{3}}\right)\left(\sqrt{\mathrm{5}+\sqrt{\mathrm{5}}}\right)\right. \\ $$

Commented by mathdave last updated on 26/Aug/20

correct

$${correct}\: \\ $$

Commented by malwan last updated on 26/Aug/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com