Exact Values of Special Angles

On the non-calculator Paper 1, the exact values of the special angles must be recalled instantly, and the answer kept exact (surds and fractions), never a decimal. This question tests that recall.

Find the exact value of $\tan 60^\circ \times \cos 30^\circ - \sin^2 30^\circ$. [3]

The working

Step 1, recall the special values: $\tan 60^\circ = \sqrt{3}, \qquad \cos 30^\circ = \frac{\sqrt{3}}{2}, \qquad \sin 30^\circ = \frac{1}{2}$

Step 2, substitute: $\tan 60^\circ \times \cos 30^\circ - \sin^2 30^\circ = \sqrt{3} \times \frac{\sqrt{3}}{2} - \left(\frac{1}{2}\right)^2 \quad \text{(M1)}$

Step 3, evaluate each part ($\sqrt{3} \times \sqrt{3} = 3$): $= \frac{3}{2} - \frac{1}{4} \quad \text{(M1)}$

Step 4, combine over a common denominator: $= \frac{6}{4} - \frac{1}{4} = \frac{5}{4} \quad \text{(A1)}$

Where the marks are won and lost

• Recall the exact values correctly, this is pure memory, and a wrong value (e.g. $\tan 60^\circ = \frac{1}{\sqrt3}$, which is actually $\tan 30^\circ$) wrecks the answer.
• $\sqrt{3} \times \frac{\sqrt{3}}{2} = \frac{3}{2}$ because $\sqrt{3} \times \sqrt{3} = 3$. Mishandling the surd product is a common slip.
• Keep everything exact, fractions, not decimals. $\frac{5}{4}$ scores; $1.25$ may lose the mark on an “exact value” question.

Common mistakes

• Confusing $\tan 30^\circ$ and $\tan 60^\circ$ (or $\sin$ and $\cos$ of $30^\circ$).
• $\sin^2 30^\circ$ meaning $(\sin 30^\circ)^2 = \frac14$, not $\sin(30^\circ)^2$.
• Converting to decimals on a non-calculator, exact-value question.

Full method: Exact Values & the Unit Circle notes. Topic home: Trigonometry pillar.