trig

Find the value of sin 2A.

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cos A = w

\(w^2+w-1=0\\ w=-0.5\pm \sqrt{0.25+1}=-0.5\pm 0.5\sqrt{5}\\ w\in \{{\color{red}-0.5(1+\sqrt{5})},\ \color{blue}-0.5(1-\sqrt{5})\}\\ sin\ 2A=2sin\ A\cdot cos\ A\\ sin\ A=\sqrt{1-cos^2\ A}\)

\(sin\ 2A=\pm 2\sqrt{1-w^2}\cdot w\\ sin\ 2A=\pm 2\sqrt{1-(-0.5(1-\sqrt{5}))^2}\cdot (-0.5(1-\sqrt{5}))\\ \color{blue}sin\ 2A\in \{0.9717,\ -0.9717\}\)

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