C3 January 2007 Q8

EdexcelOld spec6 marksProof Trigonometry

(i) Prove that\[\sec^2 x - \operatorname{cosec}^2 x \equiv \tan^2 x - \cot^2 x.\] (3)

(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 \(y\). (2)

(b) Hence evaluate \(\arccos x + \arcsin x\). Give your answer in terms of \(\pi\). (1)

Mark scheme (i)

Scheme	Marks
\(\sec^2 x - \operatorname{cosec}^2 x = (1 + \tan^2 x) - (1 + \cot^2 x)\)	M1 A1
\(= \tan^2 x - \cot^2 x\) * cso	A1
	(3)

\(\sec^2 x - \tan^2 x = 1 = \operatorname{cosec}^2 x - \cot^2 x\)	M1 A1
Rearranging \(\sec^2 x - \operatorname{cosec}^2 x = \tan^2 x - \cot^2 x\) * cso	A1 (3)

\(\left(\text{LHS} = \dfrac{1}{\cos^2 x} - \dfrac{1}{\sin^2 x} = \dfrac{\sin^2 x - \cos^2 x}{\cos^2 x\sin^2 x}\right)\)
RHS \(= \dfrac{\sin^2 x}{\cos^2 x} - \dfrac{\cos^2 x}{\sin^2 x} = \dfrac{\sin^4 x - \cos^4 x}{\cos^2 x\sin^2 x} = \dfrac{(\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x)}{\cos^2 x\sin^2 x}\)	M1
\(= \dfrac{\sin^2 x - \cos^2 x}{\cos^2 x\sin^2 x}\)	A1
\(=\) LHS * or equivalent	A1 (3)

Mark scheme (ii)(a)

Scheme	Marks
\(y = \arccos x \Rightarrow x = \cos y\)	B1
\(x = \sin\left(\dfrac{\pi}{2} - y\right) \Rightarrow \arcsin x = \dfrac{\pi}{2} - y\)	B1
	(2)

Accept \(\arcsin x = \arcsin\cos y\)