| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Proofs |
| Type | Solve equation using proven identity |
| Difficulty | Standard +0.8 This is a substantial multi-part trigonometric identity and equation question requiring compound angle formulas, algebraic manipulation, and analysis of roots. Part (i) demands careful application of tan addition formulas and simplification. Part (ii) requires converting the identity result and solving a non-trivial trigonometric equation. Part (iii) involves proving existence of roots using continuity/range arguments. The 12-mark allocation and need for proof techniques place this above average difficulty, though the methods are within C3 scope. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals |
\begin{enumerate}[label=(\roman*)]
\item Prove the identity
$$\tan(\theta + 60°) \tan(\theta - 60°) \equiv \frac{\tan^2 \theta - 3}{1 - 3 \tan^2 \theta}.$$ [4]
\item Solve, for $0° < \theta < 180°$, the equation
$$\tan(\theta + 60°) \tan(\theta - 60°) = 4 \sec^2 \theta - 3,$$
giving your answers correct to the nearest $0.1°$. [5]
\item Show that, for all values of the constant $k$, the equation
$$\tan(\theta + 60°) \tan(\theta - 60°) = k^2$$
has two roots in the interval $0° < \theta < 180°$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR C3 Q9 [12]}}