If \((\cos 35^\circ + \cos 55^\circ) = p\), then what is \(\sin 35^\circ . \cos 35^\circ\) equal to?
\(\dfrac{p^2 - 1}{2}\)
Since \(\cos 55^\circ = \cos(90^\circ - 35^\circ) = \sin 35^\circ\), we get \(p = \cos 35^\circ + \sin 35^\circ\). Squaring, \(p^2 = \cos^2 35^\circ + \sin^2 35^\circ + 2\sin 35^\circ \cos 35^\circ = 1 + 2\sin 35^\circ \cos 35^\circ\). So \(\sin 35^\circ . \cos 35^\circ = \dfrac{p^2 - 1}{2}\).
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