| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove algebraic trigonometric identity |
| Difficulty | Moderate -0.3 Part (i) is a straightforward algebraic proof using standard Pythagorean identities (sec²x = 1 + tan²x, cosec²x = 1 + cot²x), requiring only direct substitution and simplification. Part (ii) involves standard inverse trig relationships that are commonly taught and practiced. While multi-part, each component uses routine techniques without requiring novel insight or extended reasoning. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^2 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\sec^2 x - \cosec^2 x = (1+\tan^2 x) - (1+\cot^2 x) = \tan^2 x - \cot^2 x\) | M1, A1 | cso (3 marks) |
| (ii)(a) \(y = \arccos x \Rightarrow x = \cos y\); \(x = \sin\left(\frac{\pi}{2} - y\right) \Rightarrow \arcsin x = \frac{\pi}{2} - y\); Accept: arcsin \(x =\) arcsin \(\cos y\) | B1, B1 | (2 marks) |
| (b) arccos \(x +\) arcsin \(x = y + \frac{\pi}{2} - y = \frac{\pi}{2}\) | B1 | (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| Rearranging: \(\sec^2 x - \tan^2 x = 1 = \cosec^2 x - \cot^2 x\); \(\sec^2 x - \cosec^2 x = \tan^2 x - \cot^2 x\) | M1, A1 | cso (3 marks) |
| LHS \(= \frac{1}{\cos^2 x} - \frac{1}{\sin^2 x} = \frac{\sin^2 x - \cos^2 x}{\cos^2 x\sin^2 x}\); RHS \(= \frac{\sin^2 x}{\cos^2 x} - \frac{\cos^2 x}{\sin^2 x} = \frac{\sin^4 x - \cos^4 x}{\cos^2 x\sin^2 x} = \frac{(\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x)}{\cos^2 x\sin^2 x} = \frac{\sin^2 x - \cos^2 x}{\cos^2 x\sin^2 x} = \) LHS | M1, M1, A1, A1 | or equivalent (3 marks) |
(i) $\sec^2 x - \cosec^2 x = (1+\tan^2 x) - (1+\cot^2 x) = \tan^2 x - \cot^2 x$ | M1, A1 | cso (3 marks)
(ii)(a) $y = \arccos x \Rightarrow x = \cos y$; $x = \sin\left(\frac{\pi}{2} - y\right) \Rightarrow \arcsin x = \frac{\pi}{2} - y$; Accept: arcsin $x =$ arcsin $\cos y$ | B1, B1 | (2 marks)
(b) arccos $x +$ arcsin $x = y + \frac{\pi}{2} - y = \frac{\pi}{2}$ | B1 | (1 mark)
**Alternatives for (i):**
Rearranging: $\sec^2 x - \tan^2 x = 1 = \cosec^2 x - \cot^2 x$; $\sec^2 x - \cosec^2 x = \tan^2 x - \cot^2 x$ | M1, A1 | cso (3 marks)
LHS $= \frac{1}{\cos^2 x} - \frac{1}{\sin^2 x} = \frac{\sin^2 x - \cos^2 x}{\cos^2 x\sin^2 x}$; RHS $= \frac{\sin^2 x}{\cos^2 x} - \frac{\cos^2 x}{\sin^2 x} = \frac{\sin^4 x - \cos^4 x}{\cos^2 x\sin^2 x} = \frac{(\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x)}{\cos^2 x\sin^2 x} = \frac{\sin^2 x - \cos^2 x}{\cos^2 x\sin^2 x} = $ LHS | M1, M1, A1, A1 | or equivalent (3 marks)\begin{enumerate}
\item (i) Prove that
\end{enumerate}
$$\sec ^ { 2 } x - \operatorname { cosec } ^ { 2 } x \equiv \tan ^ { 2 } x - \cot ^ { 2 } x$$
(ii) Given that
$$y = \arccos x , \quad - 1 \leqslant x \leqslant 1 \text { and } 0 \leqslant y \leqslant \pi ,$$
(a) express arcsin $x$ in terms of $y$.\\
(b) Hence evaluate $\arccos x + \arcsin x$. Give your answer in terms of $\pi$.\\
\hfill \mbox{\textit{Edexcel C3 2007 Q8 [6]}}