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Исбот кунед, ки

\(\sin{A}=\sin{B}=\sin{C}=\frac{\sqrt{5}-1}{2},\)

агар \(\cos{A}=\operatorname{tg}{B}, \cos{B}=\operatorname{tg}{C}, \cos{C}=\operatorname{tg}{A}\) ва

\(A, B, C \in (0, \frac{\pi}{2})\)

\(\operatorname{tg}{\alpha}=\frac{\sin{\alpha}}{\cos{\alpha}}\)

\(\cos^2{A}=\frac{\sin^2{B}}{\cos^2{B}}\)

\(\cos^2{B}=\frac{\sin^2{C}}{\cos^2{C}}\)

\(\cos^2{C}=\frac{\sin^2{A}}{\cos^2{A}}\)

\(\cos^2{x}+\sin{x}=1\)

\(\cos^2{A}=\frac{1-\cos^2{B}}{\cos^2{B}}=\)

\(=\frac{1}{\cos^2{B}}-1=\frac{1}{\frac{\sin^2{C}}{\cos^2{C}}}-1=\)

\(=\frac{1}{\frac{1-\cos^2{C}}{\cos^2{C}}}-1=\)

\(=\frac{1}{\frac{1}{\cos^2{C}}-1}-1=\)

\(=\frac{1}{\frac{1}{\frac{1}{\cos^2{A}}-1}-1}-1\)

\(x=\cos^2{A}\)

Пас аз як чанд иваз кунии баробари мешавад:

\(\cos^2{A}=\frac{1-\cos^2{A}}{2\cdot\cos^2{A}-1}-1\)

\(x=\frac{1-x}{2\cdot x-1}-1\)

\(x\cdot(2x-1)=1-x-2x+1\)

\(2x^2-x+3x-2=0\)

\(2x^2+2x-2=0\)

\(x^2+x-1=0\)

\(D=1+4=5\)

\(x_1=\frac{-1+\sqrt{5}}{2}=\frac{\sqrt{5}-1}{2}\)

\(x_2=\frac{-1-\sqrt{5}}{2}\)

Азбаски \(0 \lt x\lt 1\),

\(x=\frac{\sqrt{5}-1}{2}\)

\(cos^2{A}=\frac{\sqrt{5}-1}{2}\)

\(sin^2{A}=1-cos^2{A}=1-\frac{\sqrt{5}-1}{2}=\)

\(=\frac{2-\sqrt{5}+1}{2}=\frac{3-\sqrt{5}}{2}\)

\(\sin{A}=\sqrt{\frac{3-\sqrt{5}}{2}}=\)

\(=\sqrt{\frac{2\cdot(3-\sqrt{5})}{2\cdot2}}=\)

\(=\sqrt{\frac{6-2\sqrt{5}}{4}}=\sqrt{\frac{5-2\sqrt{5}\cdot1+1}{4}}=\)

\(=\sqrt{\frac{(\sqrt{5}-1)^2}{2^2}}=\frac{|\sqrt{5}-1|}{2}=\frac{\sqrt{5}-1}{2}\)

\(\sin{A}=\frac{\sqrt{5}-1}{2}\)

Ҳамин тавр мебарояд, ки \(\sin{B}\) ва \(\sin{C}=\frac{\sqrt{5}-1}{2}\)

Аз инҷо мебарояд, ки

\(\sin{A}=\sin{B}=\sin{C}=\frac{\sqrt{5}-1}{2}\)

Исбот шуд.