| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Find value where max/min occurs |
| Difficulty | Standard +0.3 This is a multi-part geometry/trigonometry question that systematically builds up complexity. Parts (i)-(iii) involve standard angle-chasing and differentiation of trigonometric expressions, while part (iv) is a routine harmonic form conversion (R sin(θ-α)) commonly practiced in C3/C4. The question guides students through each step with clear instructions and 'show that' scaffolding, making it slightly easier than average despite the 16 total marks. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (i) | BAC = 120 – 90 – (90 – θ) |
| Answer | Marks |
|---|---|
| h = BC + CD = a sin θ + b sin (θ – 60°) * | B1 |
| Answer | Marks |
|---|---|
| (ii) | h = a sin θ + b sin (θ – 60°) |
| Answer | Marks |
|---|---|
| 2 2 | M1 |
| Answer | Marks |
|---|---|
| [3] | corr compound angle formula |
| Answer | Marks |
|---|---|
| (iii) | OB horizontal when h = 0 |
| Answer | Marks |
|---|---|
| 2a+b | M1 |
| Answer | Marks |
|---|---|
| [3] | sinθ |
| Answer | Marks |
|---|---|
| (iv) | 2sinθ− 3cosθ= Rsin(θ−α) |
| Answer | Marks |
|---|---|
| θ = 130.9° | M1 |
| Answer | Marks |
|---|---|
| ⇒ α = π/3 | B1 |
| Answer | Marks |
|---|---|
| [4] | R = 2 |
| Answer | Marks |
|---|---|
| = √3/4 * | B1 |
| Answer | Marks |
|---|---|
| [4] | ft their α |
Question 8:
8 | (i) | BAC = 120 – 90 – (90 – θ)
= θ – 60
BC = b sin(θ – 60)
CD = AE = a sin θ
h = BC + CD = a sin θ + b sin (θ – 60°) * | B1
M1
E1
[3]
(ii) | h = a sin θ + b sin (θ – 60°)
= a sin θ + b (sinθ cos 60 – cosθ sin 60)
= a sin θ + ½ b sin θ − √3/2 b cos θ
1 3
=(a+ b)sinθ− bcosθ*
2 2 | M1
M1
E1
[3] | corr compound angle formula
sin 60 = √3/2, cos 60 = ½
used
(iii) | OB horizontal when h = 0
1 3
(a+ b)sinθ− bcosθ=0
2 2
1 3
(a+ b)sinθ= bcosθ
2 2
3
b
sinθ 2
=
cosθ 1
a+ b
2
3b
tanθ= *
2a+b | M1
M1
E1
[3] | sinθ
=tanθ
cosθ
(iv) | 2sinθ− 3cosθ= Rsin(θ−α)
= R(sinθcosα−cosθsinα)
R cos α = 2, R sin α = √3
R2 = 22 + (√3)2 = 7, R = √7 = 2.646 m
tan α = √3/2, α = 40.9°
So h = √7 sin(θ – 40.9°)
h = √7 = 2.646 m
max
when θ – 40.9° = 90°
θ = 130.9° | M1
B1
M1A1
B1ft
M1
A1
[7]
1 3
(a+ b)sinθ= bcosθ
2 2
3
b
sinθ 2
=
cosθ 1
a+ b
2
3b
tanθ=
2a+b
2sinθ− 3cosθ= Rsin(θ−α)
= R(sinθcosα−cosθsinα)
9(i) cosθ 3 sin+θ= r cos(θ −α)
=Rcosθcosα+Rsinθsinα
⇒Rcosα=1, Rsinα= 3
⇒ R2 = 12 + (√3)2 = 4, R = 2
tan α = √3
⇒ α = π/3 | B1
M1
M1
A1
[4] | R = 2
equating correct pairs
tan α = √3 o.e.
(ii) derivative of tan θ is sec2θ
π 1 π 1 π
∫ 3 dθ=∫ 3 sec2(θ− )dθ
0 (cosθ+ 3sinθ)2 0 4 3
π
⎡1 π ⎤3
= tan(θ− )
⎢ ⎥
⎣4 3 ⎦
0
= ¼ (0 – (–√3))
= √3/4 * | B1
M1
A1
E1
[4] | ft their α
1
[tan (θ – π/3] ft their R,α(in radians)
R2
wwwIn Fig. 6, OAB is a thin bent rod, with OA = $a$ metres, AB = $b$ metres and angle OAB = 120°. The bent rod lies in a vertical plane. OA makes an angle $\theta$ above the horizontal. The vertical height BD of B above O is $h$ metres. The horizontal through A meets BD at C and the vertical through A meets OD at E.
\includegraphics{figure_6}
\begin{enumerate}[label=(\roman*)]
\item Find angle BAC in terms of $\theta$. Hence show that
$$h = a\sin\theta + b\sin(\theta - 60°).$$ [3]
\item Hence show that $h = (a + \frac{1}{2}b)\sin\theta - \frac{\sqrt{3}}{2}b\cos\theta$. [3]
\end{enumerate}
The rod now rotates about O, so that $\theta$ varies. You may assume that the formulae for $h$ in parts (i) and (ii) remain valid.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Show that OB is horizontal when $\tan\theta = \frac{\sqrt{3}b}{2a + b}$. [3]
\end{enumerate}
In the case when $a = 1$ and $b = 2$, $h = 2\sin\theta - \sqrt{3}\cos\theta$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{3}
\item Express $2\sin\theta - \sqrt{3}\cos\theta$ in the form $R\sin(\theta - \alpha)$. Hence, for this case, write down the maximum value of $h$ and the corresponding value of $\theta$. [7]
\end{enumerate}
\hfill \mbox{\textit{OCR C4 Q8 [16]}}