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how did they get \[1/\sqrt{2}\]?
Their equation is \[16\sqrt{2}sin\theta=8+8\sqrt2sin\theta\]We start by subtracting 8sqrt{2}sin(theta) on both sides: \[16\sqrt{2}sin\theta - 8\sqrt2sin\theta=8+8\sqrt2sin\theta - 8\sqrt2sin\theta\]\[8\sqrt2sin\theta=8\]\[\frac{8\sqrt2sin\theta}{8\sqrt2}=\frac{8}{8\sqrt2}\]\[sin\theta=\frac{1}{\sqrt2}\]
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