Work out the exact value of \(\sin 60^\circ \times \cos 30^\circ\), giving your answer as a fraction in its simplest form.

Both values come from the half of an equilateral triangle, the right-angled triangle with sides 1, \(\sqrt{3}\) and 2.

1. \(\sin 60^{\circ}=\dfrac{\sqrt{3}}{2}\), the side opposite \(60^{\circ}\) over the hypotenuse.
2. \(\cos 30^{\circ}=\dfrac{\sqrt{3}}{2}\), the side adjacent to \(30^{\circ}\) over the hypotenuse.

These are equal because the two angles are complementary: the side opposite one is the side adjacent to the other. So the product is

\(\dfrac{\sqrt{3}}{2}\times\dfrac{\sqrt{3}}{2}\) [M1].

Multiply numerators and denominators: \(\sqrt{3}\times\sqrt{3}=3\) and \(2\times 2=4\), giving

\(\dfrac{3}{4}\) [A1] cao.

The step worth fixing is \(\sqrt{3}\times\sqrt{3}=3\). Squaring a square root returns the original number, so the surd disappears and the answer is a plain fraction. Writing \(\sqrt{9}\) or \(\sqrt{6}\) here are the two common errors.