OCR C3 — Question 7 11 marks

Exam BoardOCR
ModuleC3 (Core Mathematics 3)
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReciprocal Trig & Identities
TypeProve identity then solve equation
DifficultyStandard +0.3 This is a standard C3 trigonometric identities question with three routine parts: (i) derive a double-angle formula using a given identity (straightforward substitution), (ii) prove a standard half-angle identity using part (i), and (iii) solve an equation by substituting the proven identities. All techniques are textbook exercises requiring methodical application rather than 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

7. (i) Use the identity $$\cos ( A + B ) \equiv \cos A \cos B - \sin A \sin B$$ to prove that $$\cos x \equiv 1 - 2 \sin ^ { 2 } \frac { x } { 2 }$$ (ii) Prove that, for \(\sin x \neq 0\), $$\frac { 1 - \cos x } { \sin x } \equiv \tan \frac { x } { 2 }$$ (iii) Find the values of \(x\) in the interval \(0 \leq x \leq 360 ^ { \circ }\) for which $$\frac { 1 - \cos x } { \sin x } = 2 \sec ^ { 2 } \frac { x } { 2 } - 5$$ giving your answers to 1 decimal place where appropriate.

Question 7:
Part (i)
AnswerMarks Guidance
AnswerMark Notes
\(\cos(A+B) \equiv \cos A\cos B - \sin A\sin B\)
Let \(A = B = \frac{x}{2}\): \(\cos x \equiv \cos^2\frac{x}{2} - \sin^2\frac{x}{2}\)M1
\(\cos x \equiv (1 - \sin^2\frac{x}{2}) - \sin^2\frac{x}{2}\)
\(\cos x \equiv 1 - 2\sin^2\frac{x}{2}\)A1
Part (ii)
AnswerMarks Guidance
AnswerMark Notes
\(\text{LHS} \equiv \frac{1-(1-2\sin^2\frac{x}{2})}{2\sin\frac{x}{2}\cos\frac{x}{2}}\)M1
\(\equiv \frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} \equiv \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}\)M1
\(\equiv \tan\frac{x}{2} \equiv \text{RHS}\)A1
Part (iii)
AnswerMarks Guidance
AnswerMark Notes
\(\tan\frac{x}{2} = 2\sec^2\frac{x}{2} - 5,\quad \tan\frac{x}{2} = 2(1+\tan^2\frac{x}{2}) - 5\)M1
\(2\tan^2\frac{x}{2} - \tan\frac{x}{2} - 3 = 0,\quad (2\tan\frac{x}{2}-3)(\tan\frac{x}{2}+1) = 0\)M1
\(\tan\frac{x}{2} = -1\) or \(\frac{3}{2}\)A1
\(\frac{x}{2} = 135°\) or \(56.310°\)M1
\(x = 112.6°\) (1dp), \(270°\)A2 (11)
# Question 7:

## Part (i)
| Answer | Mark | Notes |
|--------|------|-------|
| $\cos(A+B) \equiv \cos A\cos B - \sin A\sin B$ | | |
| Let $A = B = \frac{x}{2}$: $\cos x \equiv \cos^2\frac{x}{2} - \sin^2\frac{x}{2}$ | M1 | |
| $\cos x \equiv (1 - \sin^2\frac{x}{2}) - \sin^2\frac{x}{2}$ | | |
| $\cos x \equiv 1 - 2\sin^2\frac{x}{2}$ | A1 | |

## Part (ii)
| Answer | Mark | Notes |
|--------|------|-------|
| $\text{LHS} \equiv \frac{1-(1-2\sin^2\frac{x}{2})}{2\sin\frac{x}{2}\cos\frac{x}{2}}$ | M1 | |
| $\equiv \frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} \equiv \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}$ | M1 | |
| $\equiv \tan\frac{x}{2} \equiv \text{RHS}$ | A1 | |

## Part (iii)
| Answer | Mark | Notes |
|--------|------|-------|
| $\tan\frac{x}{2} = 2\sec^2\frac{x}{2} - 5,\quad \tan\frac{x}{2} = 2(1+\tan^2\frac{x}{2}) - 5$ | M1 | |
| $2\tan^2\frac{x}{2} - \tan\frac{x}{2} - 3 = 0,\quad (2\tan\frac{x}{2}-3)(\tan\frac{x}{2}+1) = 0$ | M1 | |
| $\tan\frac{x}{2} = -1$ or $\frac{3}{2}$ | A1 | |
| $\frac{x}{2} = 135°$ or $56.310°$ | M1 | |
| $x = 112.6°$ (1dp), $270°$ | A2 | **(11)** |

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7. (i) Use the identity

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

to prove that

$$\cos x \equiv 1 - 2 \sin ^ { 2 } \frac { x } { 2 }$$

(ii) Prove that, for $\sin x \neq 0$,

$$\frac { 1 - \cos x } { \sin x } \equiv \tan \frac { x } { 2 }$$

(iii) Find the values of $x$ in the interval $0 \leq x \leq 360 ^ { \circ }$ for which

$$\frac { 1 - \cos x } { \sin x } = 2 \sec ^ { 2 } \frac { x } { 2 } - 5$$

giving your answers to 1 decimal place where appropriate.\\

\hfill \mbox{\textit{OCR C3  Q7 [11]}}