Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 144959 by mathdanisur last updated on 30/Jun/21

Answered by Ar Brandon last updated on 01/Jul/21

$$\sin 72°=\cos 18°=\frac{\sqrt{10+2\sqrt{5}}}{4}$$

Commented by Ar Brandon last updated on 01/Jul/21

$${\cos }^{2}18°+{\sin }^{2}18°=1$$$$\cos 18°=\pm \sqrt{1-{\sin }^{2}18°}=\pm \sqrt{1-{\left(\frac{\sqrt{5}-1}{4}\right)}^2}$$$$=\pm \sqrt{\frac{10+2\sqrt{5}}{16}}=\frac{1}{4}\sqrt{10+2\sqrt{5}}\{\because \cos \theta >0\mathrm{\forall }\theta \in (0,90°)\}$$

Commented by mathdanisur last updated on 01/Jul/21

$$coolthankyouSir$$

Answered by imjagoll last updated on 01/Jul/21

$$\sin 72°=\sin (90°-18°)=\cos 18°$$$$=\sqrt{1-\sin {}^{2}18°}$$$$=\sqrt{1-\left(\frac{6-2\sqrt{5}}{16}\right)}=\sqrt{\frac{10+2\sqrt{5}}{16}}$$$$=\frac{1}{4}\sqrt{10+2\sqrt{5}};\mathrm{x}=4$$

Commented by mathdanisur last updated on 01/Jul/21

$$coolthanksSir$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com