Question

Это можно в человеческий вид ? 8 + sin(36) * 8 / sin (72) cos (36) * 8 + tg (72) / (sin(36) * 8)

Answer 1

По формуле двойного угла

sin(2a)=2*sina*cosa

$8+\frac{8\sin 36^0}{\sin 72^0}=8+\frac{8\sin 36^0}{2\sin 36^0\cos 36^0}$

$8+\frac{8\sin 36^0}{2\sin 36^0\cos 36^0}=8+\frac{4}{\cos 36^0}$

Известно, что $\cos36^0=\frac{1+\sqrt{5}}{4}$

$8+\frac{4}{\cos 36^0}=8+\frac{4}{\frac{1+\sqrt{5}}{4}}$

$8+\frac{4}{\frac{1+\sqrt{5}}{4}}=8+\frac{16}{1+\sqrt{5}}$

$8+\frac{16}{1+\sqrt{5}}=8+\frac{16*(\sqrt{5}-1)}{(1+\sqrt{5})*(\sqrt{5}-1)}$

$8+\frac{16*(\sqrt{5}-1)}{(1+\sqrt{5})*(\sqrt{5}-1)}=8+\frac{16*(\sqrt{5}-1)}{(\sqrt{5})^2-1^2}$

$8+\frac{16*(\sqrt{5}-1)}{(\sqrt{5})^2-1^2}=8+\frac{16*(\sqrt{5}-1)}{4}$

$8+\frac{16*(\sqrt{5}-1)}{4}=8+4(\sqrt{5}-1)$

$8+4(\sqrt{5}-1)=4+4\sqrt{5}$

$4+4\sqrt{5}=4*(1+\sqrt{5})$

Во втором примере воспользуемся формулой

$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}$

$8*\cos 36^0+\frac{\tan 72^0}{8*\sin 36^0}=8*\frac{\sqrt{5}+1}{4}+\frac{\frac{\sin 72^0}{\cos 72^0}}{8*\sin 36^0}$

$8*\frac{\sqrt{5}+1}{4}+\frac{\frac{\sin 72^0}{\cos 72^0}}{8*\sin 36^0}=2*(\sqrt{5}+1)+\frac{\sin 72^0}{8\sin 36^0\cos 72^0}$

$2*(\sqrt{5}+1)+\frac{\sin 72^0}{8\sin 36^0\cos 72^0}=2*(\sqrt{5}+1)+\frac{2\sin 36^0\cos 36^0}{8\sin 36^0\cos 72^0}$

$2*(\sqrt{5}+1)+\frac{2\sin 36^0\cos 36^0}{8\sin 36^0\cos 72^0}=2*(\sqrt{5}+1)+\frac{\cos 36^0}{4\cos 72^0}\quad(2)$

Вычислим отдельно $\cos 72^0=\cos(2*36^0)$

По формуле двойного угла для косинуса

$\cos2\alpha=2\cos^2\alpha-1$

$\cos(2*36^0)=2\cos^2 36^0-1$

$2\cos^2 36^0-1=2*\left(\frac{\sqrt{5}+1}{4}\right)^2-1$

$2*\left(\frac{\sqrt{5}+1}{4}\right)^2-1=\frac{(\sqrt{5}+1)^2}{8}-1$

$\frac{(\sqrt{5}+1)^2}{8}-1=\frac{6+2\sqrt{5}}{8}-1$

$\frac{6+2\sqrt{5}}{8}-1=\frac{3+\sqrt{5}}{4}-1$

$\frac{3+\sqrt{5}}{4}-1=\frac{3+\sqrt{5}-4}{4}$

$\frac{3+\sqrt{5}-4}{4}=\frac{\sqrt{5}-1}{4}$

Значит

$\cos 72^0=\frac{\sqrt{5}-1}{4}$

Вернемся к (2)

$2*(\sqrt{5}+1)+\frac{\cos 36^0}{4\cos 72^0}=2*(\sqrt{5}+1)+\frac{\cos 36^0}{\sqrt{5}-1}$

$2*(\sqrt{5}+1)+\frac{\cos 36^0}{\sqrt{5}-1}=2*(\sqrt{5}+1)+\frac{\sqrt{5}+1}{4(\sqrt{5}-1)}$

$2*(\sqrt{5}+1)+\frac{\sqrt{5}+1}{4(\sqrt{5}-1)}=2*(\sqrt{5}+1)+\frac{(\sqrt{5}+1)^2}{4(\sqrt{5}-1)*(\sqrt{5}+1)}$

$2*(\sqrt{5}+1)+\frac{(\sqrt{5}+1)^2}{4(\sqrt{5}-1)*(\sqrt{5}+1)}=2*(\sqrt{5}+1)+\frac{(\sqrt{5}+1)^2}{4*4}$

$2*(\sqrt{5}+1)+\frac{(\sqrt{5}+1)^2}{4*4}=2*(\sqrt{5}+1)+\frac{(6+2\sqrt{5})}{4*4}$

$2*(\sqrt{5}+1)+\frac{(6+2\sqrt{5})}{4*4}=2*(\sqrt{5}+1)+\frac{(3+\sqrt{5})}{8}$

$2*(\sqrt{5}+1)+\frac{(3+\sqrt{5})}{8}=\frac{(16+3+16\sqrt{5}+\sqrt{5})}{8}$

$\frac{(16+3+16\sqrt{5}+\sqrt{5})}{8}=\frac{(19+17\sqrt{5})}{8}$