OCR C3 2014 June — Question 2 6 marks

Exam BoardOCR
ModuleC3 (Core Mathematics 3)
Year2014
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAddition & Double Angle Formulae
TypeSolve equation with sec/cosec/cot
DifficultyStandard +0.3 This question requires knowing the double angle formula (cos 2θ = 1 - 2sin²θ) and that cosec θ = 1/sin θ, then solving a quadratic in sin θ. It's a standard C3 exercise with clear signposting ('by first using appropriate identities') and straightforward algebraic manipulation, making it slightly easier than average.
Spec1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals

2 By first using appropriate identities, solve the equation $$5 \cos 2 \theta \operatorname { cosec } \theta = 2$$ for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
State or imply \(\cosec\theta = 1 \div \sin\theta\)B1 allow \(\cosec = 1 \div \sin\)
Attempt to express equation in terms of \(\sin\theta\) onlyM1 using identity of form \(\pm 1 \pm 2\sin^2\theta\) for \(\cos 2\theta\)
Obtain \(10\sin^2\theta + 2\sin\theta - 5 = 0\)A1 or unsimplified equiv involving \(\sin\theta\) only but with no \(\sin\theta\) remaining in denominator
Attempt use of formula to find \(\sin\theta\) from 3-term quadratic equation involving \(\sin\theta\) (using formula or completing square even if their equation can be solved by factorisation)M1 use implied by at least one correct value of \(\sin\theta\) or \(\theta\); if correct quadratic formula quoted, condone one sign error for M1; if formula not first quoted, any error leads to M0. If completion of square used to solve equation, this must be correct for M1 to be earned
Obtain \(37.9°\)A1 or greater accuracy \(37.8896\ldots\)
Obtain \(142°\)A1 or greater accuracy \(142.1103\ldots\); and no others between 0 and 180; ignore any answers outside \(0\)–\(180\). No working and answers only (max 2/6): 37.9 (or greater accuracy) B1; 142 (or greater accuracy) and no others … B1
Total[6]
## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply $\cosec\theta = 1 \div \sin\theta$ | B1 | allow $\cosec = 1 \div \sin$ |
| Attempt to express equation in terms of $\sin\theta$ only | M1 | using identity of form $\pm 1 \pm 2\sin^2\theta$ for $\cos 2\theta$ |
| Obtain $10\sin^2\theta + 2\sin\theta - 5 = 0$ | A1 | or unsimplified equiv involving $\sin\theta$ only but with no $\sin\theta$ remaining in denominator |
| Attempt use of formula to find $\sin\theta$ from 3-term quadratic equation involving $\sin\theta$ (using formula or completing square even if their equation can be solved by factorisation) | M1 | use implied by at least one correct value of $\sin\theta$ or $\theta$; if correct quadratic formula quoted, condone one sign error for M1; if formula not first quoted, any error leads to M0. If completion of square used to solve equation, this must be correct for M1 to be earned |
| Obtain $37.9°$ | A1 | or greater accuracy $37.8896\ldots$ |
| Obtain $142°$ | A1 | or greater accuracy $142.1103\ldots$; and no others between 0 and 180; ignore any answers outside $0$–$180$. No working and answers only (max 2/6): 37.9 (or greater accuracy) B1; 142 (or greater accuracy) and no others … B1 |
| **Total** | **[6]** | |

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2 By first using appropriate identities, solve the equation

$$5 \cos 2 \theta \operatorname { cosec } \theta = 2$$

for $0 ^ { \circ } < \theta < 180 ^ { \circ }$.

\hfill \mbox{\textit{OCR C3 2014 Q2 [6]}}