Edexcel C3 2008 January — Question 6 11 marks

Exam BoardEdexcel
ModuleC3 (Core Mathematics 3)
Year2008
SessionJanuary
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReciprocal Trig & Identities
TypeProve identity then solve equation
DifficultyStandard +0.3 This is a standard C3 question with routine techniques: part (a) uses double angle formulae mechanically to derive cos 3x; part (b)(i) is algebraic manipulation of a trig identity using common denominators; part (b)(ii) applies the proven identity to solve a straightforward equation. All steps are textbook exercises requiring no novel insight, making it slightly easier than average.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals

6. (a) Use the double angle formulae and the identity $$\cos ( A + B ) \equiv \cos A \cos B - \sin A \sin B$$ to obtain an expression for \(\cos 3 x\) in terms of powers of \(\cos x\) only.
(b) (i) Prove that $$\frac { \cos x } { 1 + \sin x } + \frac { 1 + \sin x } { \cos x } \equiv 2 \sec x , \quad x \neq ( 2 n + 1 ) \frac { \pi } { 2 }$$ (ii) Hence find, for \(0 < x < 2 \pi\), all the solutions of $$\frac { \cos x } { 1 + \sin x } + \frac { 1 + \sin x } { \cos x } = 4$$

Question 6:
Part (a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\cos(2x+x) = \cos 2x\cos x - \sin 2x\sin x\)M1
\(= (2\cos^2 x - 1)\cos x - (2\sin x\cos x)\sin x\)M1
\(= (2\cos^2 x -1)\cos x - 2(1-\cos^2 x)\cos x\)A1 any correct expression
\(= 4\cos^3 x - 3\cos x\)A1
Part (b)(i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{\cos x}{1+\sin x} + \frac{1+\sin x}{\cos x} = \frac{\cos^2 x + (1+\sin x)^2}{(1+\sin x)\cos x}\)M1
\(= \frac{\cos^2 x + 1 + 2\sin x + \sin^2 x}{(1+\sin x)\cos x}\)A1
\(= \frac{2(1+\sin x)}{(1+\sin x)\cos x}\)M1
\(= \frac{2}{\cos x} = 2\sec x\)A1 cso
Part (c):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\sec x = 2\) or \(\cos x = \frac{1}{2}\)M1
\(x = \frac{\pi}{3}, \frac{5\pi}{3}\)A1, A1 accept awrt 1.05, 5.24
# Question 6:

## Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\cos(2x+x) = \cos 2x\cos x - \sin 2x\sin x$ | M1 | |
| $= (2\cos^2 x - 1)\cos x - (2\sin x\cos x)\sin x$ | M1 | |
| $= (2\cos^2 x -1)\cos x - 2(1-\cos^2 x)\cos x$ | A1 | any correct expression |
| $= 4\cos^3 x - 3\cos x$ | A1 | |

## Part (b)(i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{\cos x}{1+\sin x} + \frac{1+\sin x}{\cos x} = \frac{\cos^2 x + (1+\sin x)^2}{(1+\sin x)\cos x}$ | M1 | |
| $= \frac{\cos^2 x + 1 + 2\sin x + \sin^2 x}{(1+\sin x)\cos x}$ | A1 | |
| $= \frac{2(1+\sin x)}{(1+\sin x)\cos x}$ | M1 | |
| $= \frac{2}{\cos x} = 2\sec x$ | A1 | cso |

## Part (c):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sec x = 2$ or $\cos x = \frac{1}{2}$ | M1 | |
| $x = \frac{\pi}{3}, \frac{5\pi}{3}$ | A1, A1 | accept awrt 1.05, 5.24 |

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6. (a) Use the double angle formulae and the identity

$$\cos ( A + B ) \equiv \cos A \cos B - \sin A \sin B$$

to obtain an expression for $\cos 3 x$ in terms of powers of $\cos x$ only.\\
(b) (i) Prove that

$$\frac { \cos x } { 1 + \sin x } + \frac { 1 + \sin x } { \cos x } \equiv 2 \sec x , \quad x \neq ( 2 n + 1 ) \frac { \pi } { 2 }$$

(ii) Hence find, for $0 < x < 2 \pi$, all the solutions of

$$\frac { \cos x } { 1 + \sin x } + \frac { 1 + \sin x } { \cos x } = 4$$

\hfill \mbox{\textit{Edexcel C3 2008 Q6 [11]}}