OCR MEI C2 2014 June — Question 8 3 marks

Exam BoardOCR MEI
ModuleC2 (Core Mathematics 2)
Year2014
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard trigonometric equations
TypeFunction properties and inverses
DifficultyEasy -1.2 This is a straightforward trigonometric identity question requiring only basic recall: recognizing that √(1-cos²θ) = sin θ (using the Pythagorean identity) and that sin θ/tan θ = cos θ. It's a routine 3-mark question with no problem-solving or novel insight required, making it easier than average.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1

Simplify \(\frac{\sqrt{1 - \cos^2 \theta}}{\tan \theta}\), where \(\theta\) is an acute angle. [3]

AnswerMarks Guidance
\(\frac{\sqrt{\sin^2\theta}}{\sin\theta}\) or \(\frac{\cos\theta\sqrt{\sin^2\theta}}{\sin\theta}\)M1 correct substitution for numerator; allow maximum of M1M1 if \(\pm\sqrt{\sin^2\theta}\) oe substituted
\(\cos\theta\)M1 correct substitution for denominator
\(\cos\theta\) caoA1 A0 if follows wrong working or B3 www or if unsupported; mark the final answer but ignore attempts to solve for \(\theta\); allow recovery from omission of \(\theta\)
Total: [3]
$\frac{\sqrt{\sin^2\theta}}{\sin\theta}$ or $\frac{\cos\theta\sqrt{\sin^2\theta}}{\sin\theta}$ | M1 | correct substitution for numerator; allow maximum of M1M1 if $\pm\sqrt{\sin^2\theta}$ oe substituted
$\cos\theta$ | M1 | correct substitution for denominator
$\cos\theta$ cao | A1 | A0 if follows wrong working or B3 www or if unsupported; mark the final answer but ignore attempts to solve for $\theta$; allow recovery from omission of $\theta$
**Total: [3]**

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Simplify $\frac{\sqrt{1 - \cos^2 \theta}}{\tan \theta}$, where $\theta$ is an acute angle. [3]

\hfill \mbox{\textit{OCR MEI C2 2014 Q8 [3]}}