What is the value of \(\frac{3}{2}\left( {\frac{{\cos 39}}{{\sin 51}}} \right) - \sqrt {({{\sin }^2}39 + {{\sin }^2}51})\) ?
either 1/2 or 5/2
Since \(51^\circ=90^\circ-39^\circ\), we have \(\sin 51^\circ=\cos 39^\circ\), so the first term reduces:
\[\frac{3}{2}\cdot\frac{\cos 39^\circ}{\sin 51^\circ}=\frac{3}{2}\cdot 1=\frac{3}{2}\]
Under the root, \(\sin^2 39^\circ+\sin^2 51^\circ=\sin^2 39^\circ+\cos^2 39^\circ=1\), so \(\sqrt{1}=\pm1\). The expression becomes \(\tfrac{3}{2}\mp1\), giving \(\tfrac{1}{2}\) or \(\tfrac{5}{2}\).
Hence the answer is either 1/2 or 5/2.
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