OCR C3 2009 June — Question 7 10 marks

Exam BoardOCR
ModuleC3 (Core Mathematics 3)
Year2009
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHarmonic Form
TypeFind maximum or minimum value
DifficultyStandard +0.3 This is a standard C3 harmonic form question with routine application. Part (i) uses the textbook R sin(θ-α) method with straightforward arithmetic (R=10, α≈36.9°). Part (ii)(a) requires solving sin(θ-α)=0.9 and finding two solutions in range. Part (ii)(b) involves recognizing that the expression is 2[R₁sin(x-α)] - [R₂sin(y-α)] and maximizing by setting sin terms to ±1, which is slightly above routine but still a standard technique. The multi-part structure and 10 marks indicate moderate length, but no novel insight is required—just careful execution of learned methods.
Spec1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals

  1. Express \(8 \sin \theta - 6 \cos \theta\) in the form \(R \sin(\theta - \alpha)\), where \(R > 0\) and \(0° < \alpha < 90°\). [3]
  2. Hence
    1. solve, for \(0° < \theta < 360°\), the equation \(8 \sin \theta - 6 \cos \theta = 9\), [4]
    2. find the greatest possible value of $$32 \sin x - 24 \cos x - (16 \sin y - 12 \cos y)$$ as the angles \(x\) and \(y\) vary. [3]

Part (i)
AnswerMarks Guidance
State \(R = 10\)B1 or equiv
Attempt to find value of \(\alpha\)M1 implied by correct answer or its complement; allow sin/cos muddles
Obtain \(36.9\) or \(\tan^{-1}\frac{3}{4}\)A1\3 or greater accuracy 36.8699…
Part (ii)(a)
AnswerMarks Guidance
Show correct process for finding one angleM1
Obtain (64.16 + 36.87 and hence) 101A1
Show correct process for finding second angleM1
Obtain (115.84 + 36.87 and hence) 153A1\4 following their value of \(\alpha\); or greater accuracy 152.711…; and no other between 0 and 360
Part (ii)(b)
AnswerMarks Guidance
Recognise link with part (i)M1 signalled by 40 … – 20 …
Use fact that maximum and minimum values of sine are 1 and –1M1 may be implied; or equiv
Obtain 60A1
## Part (i)

State $R = 10$ | B1 | or equiv
Attempt to find value of $\alpha$ | M1 | implied by correct answer or its complement; allow sin/cos muddles

Obtain $36.9$ or $\tan^{-1}\frac{3}{4}$ | A1\3 | or greater accuracy 36.8699…

## Part (ii)(a)

Show correct process for finding one angle | M1 |
Obtain (64.16 + 36.87 and hence) 101 | A1 |
Show correct process for finding second angle | M1 |
Obtain (115.84 + 36.87 and hence) 153 | A1\4 | following their value of $\alpha$; or greater accuracy 152.711…; and no other between 0 and 360

## Part (ii)(b)

Recognise link with part (i) | M1 | signalled by 40 … – 20 …
Use fact that maximum and minimum values of sine are 1 and –1 | M1 | may be implied; or equiv
Obtain 60 | A1 | 

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\begin{enumerate}[label=(\roman*)]
\item Express $8 \sin \theta - 6 \cos \theta$ in the form $R \sin(\theta - \alpha)$, where $R > 0$ and $0° < \alpha < 90°$. [3]
\item Hence
\begin{enumerate}[label=(\alph*)]
\item solve, for $0° < \theta < 360°$, the equation $8 \sin \theta - 6 \cos \theta = 9$, [4]
\item find the greatest possible value of
$$32 \sin x - 24 \cos x - (16 \sin y - 12 \cos y)$$
as the angles $x$ and $y$ vary. [3]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{OCR C3 2009 Q7 [10]}}