2026-08-31T11:34:12.100823 image/svg+xml Matplotlib v3.10.9, https://matplotlib.org/ © EAGLE BEACON GLOBAL Figure 6 shows the triangle \(ABC\), in which \(A...

The purpose of part (a) is to produce an exact value for \(\cos 15^\circ\), which no calculator display can give in surd form. The route is to write \(15^\circ\) as a difference of two angles whose sine and cosine are known exactly, and \(45^\circ - 30^\circ\) is the natural choice. Part (b) then feeds that exact value into the cosine rule, which is the correct rule here because two sides and the angle between them are known and the side opposite that angle is wanted; the sine rule cannot be started, since it would need a side and its opposite angle as a matched pair.

(a) Show that \(\cos 15^\circ = \dfrac{\sqrt{6} + \sqrt{2}}{4}\) [3]

The result is printed, so the marks are for the derivation. Quote the addition formula:

\[\cos(A - B) = \cos A\cos B + \sin A\sin B \qquad \textbf{[M1]}\]

The [M1] is for the correct formula, and the sign is the thing to get right: for \(\cos(A - B)\) the connective is \(+\), the opposite of the sign in the bracket. Now substitute \(A = 45^\circ\) and \(B = 30^\circ\), as the question directs:

\[\cos 15^\circ = \cos 45^\circ\cos 30^\circ + \sin 45^\circ\sin 30^\circ \qquad \textbf{[M1]}\]

The second [M1] is for this substitution. Insert the exact values \(\cos 45^\circ = \sin 45^\circ = \dfrac{\sqrt{2}}{2}\), \(\cos 30^\circ = \dfrac{\sqrt{3}}{2}\) and \(\sin 30^\circ = \dfrac{1}{2}\):

\[= \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4} \qquad \textbf{[A1]}\]

The [A1] is for reaching the printed form. Note \(\sqrt{2} \times \sqrt{3} = \sqrt{6}\), using \(\sqrt{m}\sqrt{n} = \sqrt{mn}\). The wrong turn this part is built to catch is writing \(\cos 15^\circ = \cos 45^\circ - \cos 30^\circ\); the cosine of a difference is not the difference of the cosines, and a quick decimal test exposes it, since \(0.7071 - 0.8660 = -0.1589\) while \(\cos 15^\circ \approx 0.9659\).

(b) Hence find the exact value of \(BC^2\) [3]

In the triangle \(ABC\) shown, \(AB = 4\) cm and \(AC = 6\) cm are the two sides enclosing the known angle \(BAC = 15^\circ\), and \(BC\) is opposite it. That is precisely the configuration the cosine rule handles:

\[BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(BAC) \qquad \textbf{[M1]}\]

The [M1] is for a correct statement of the cosine rule with the right angle paired to the right side. Substituting, and using the exact value from part (a) rather than a decimal:

\[BC^2 = 4^2 + 6^2 - 2 \times 4 \times 6 \times \frac{\sqrt{6} + \sqrt{2}}{4} = 16 + 36 - 48 \times \frac{\sqrt{6} + \sqrt{2}}{4} \qquad \textbf{[M1]}\] \[= 52 - 12\left(\sqrt{6} + \sqrt{2}\right) = 52 - 12\sqrt{6} - 12\sqrt{2} \qquad \textbf{[A1]}\]

so \(p = 52\), \(q = -12\) and \(r = -12\). The second [M1] is for the substitution, and it is a follow-through mark: a candidate whose part (a) went wrong can still earn it by substituting their own surd correctly. The [A1] needs the three integers, and \(48 \div 4 = 12\) is the arithmetic that must not slip.

Two errors are common. The first is switching to a decimal for \(\cos 15^\circ\), which gives \(BC^2 \approx 5.63\) and cannot be written in the required form \(p + q\sqrt{6} + r\sqrt{2}\), so the accuracy mark is lost however correct the number is; the word "exact" forbids it. The second is failing to distribute the minus sign across the bracket, producing \(52 - 12\sqrt{6} + 12\sqrt{2}\). Note also that the question asks for \(BC^2\), not \(BC\), so no square root should be taken at the end.

Check: evaluate both forms numerically. \(52 - 12(2.44949) - 12(1.41421) = 52 - 29.394 - 16.971 = 5.635\), and directly \(16 + 36 - 48\cos 15^\circ = 52 - 46.365 = 5.635\). They agree, so the surd expression is right. A magnitude check also reassures: \(BC = \sqrt{5.635} \approx 2.37\) cm, which is sensibly small because the \(15^\circ\) angle is narrow, and it satisfies the triangle inequality with sides \(4\) and \(6\), since \(6 - 4 = 2 \lt 2.37 \lt 10\).