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Question Number 189125 by Rupesh123 last updated on 12/Mar/23

Answered by cortano12 last updated on 12/Mar/23

$$\cos 18°={2\cos }^{2}9°-1$$$${\cos }^{2}9°=\frac{1+\cos 18°}{2}$$$$\cos 18°=\sqrt{1-\sin {}^{2}18°}$$$$=\sqrt{1-{\left(\frac{\sqrt{5}-1}{4}\right)}^2}$$$$=\frac{\sqrt{16-(6-2\sqrt{5})}}{4}=\frac{\sqrt{10+2\sqrt{5}}}{4}$$$$$$

Commented by Rupesh123 last updated on 13/Mar/23

Excellent!

Answered by BaliramKumar last updated on 12/Mar/23

$$$$$$cos18°=\frac{\sqrt{10+2\sqrt{5}}}{4}$$$$2{cos}^{2}9°-1=\frac{\sqrt{10+2\sqrt{5}}}{4}$$$$8{cos}^{2}9°-4=\sqrt{10+2\sqrt{5}}$$$$8{cos}^{2}9°=4+\sqrt{10+2\sqrt{5}}$$$$2\sqrt{2}cos9°=\sqrt{4+\sqrt{10+2\sqrt{5}}}$$$$2\sqrt{2}cos9°=\sqrt{\frac{4+\sqrt{16-10-2\sqrt{5}}}{2}}+\sqrt{\frac{4-\sqrt{16-10-2\sqrt{5}}}{2}}$$$$2\sqrt{2}cos9°=\sqrt{\frac{4+\sqrt{6-2\sqrt{5}}}{2}}+\sqrt{\frac{4-\sqrt{6-2\sqrt{5}}}{2}}$$$$2\sqrt{2}cos9°=\sqrt{\frac{4+\sqrt{5}-1}{2}}+\sqrt{\frac{4-(\sqrt{5}-1)}{2}}$$$$2\sqrt{2}cos9°=\sqrt{\frac{3+\sqrt{5}}{2}}+\sqrt{\frac{5-\sqrt{5}}{2}}$$$$cos9°=\frac{\sqrt{3+\sqrt{5}}}{4}+\frac{\sqrt{5-\sqrt{5}}}{4}=\frac{2\sqrt{3+\sqrt{5}}+2\sqrt{5-\sqrt{5}}}{8}$$$$cos9°=\frac{\sqrt{2}\sqrt{6+2\sqrt{5}}+2\sqrt{5-\sqrt{5}}}{8}$$$$cos9°=\frac{\sqrt{2}(1+\sqrt{5})+2\sqrt{5-\sqrt{5}}}{8}$$$$cos9°=\frac{\sqrt{2}+\sqrt{10}+2\sqrt{5-\sqrt{5}}}{8}$$$$$$$$$$$$\sqrt{a\pm \sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\pm \sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$$$$$

Commented by Rupesh123 last updated on 13/Mar/23

Excellent!

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