| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2021 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Prove identity with double/compound angles |
| Difficulty | Moderate -0.3 Part (a) is a guided geometric proof of the sine addition formula with scaffolded steps requiring basic trigonometry (SOHCAHTOA) and triangle area formulas. Part (b) involves routine application of compound angle formulae and algebraic manipulation. While multi-step, each component is standard A-level technique with no novel insight required, making it slightly easier than average. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C)1.05l Double angle formulae: and compound angle formulae1.05m Geometric proofs: of trig sum and double angle formulae1.05p Proof involving trig: functions and identities |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos y = \frac{CD}{a}\), hence \(CD = a\cos y\) | B1 | Justification for \(CD\). Need to see either \(\cos y = \frac{CD}{a}\) or \(\text{adj} = \text{hyp} \times \cos\theta\) before given answer. |
| Answer | Marks | Guidance |
|---|---|---|
| Area \(= \frac{1}{2}AC \cdot CD\sin x = \frac{1}{2}b(a\cos y)\sin x = \frac{1}{2}ab\sin x\cos y\) A.G. | B1 | Use area of triangle to show given answer. Could quote general expression for area and then show clear substitution. Condone not being rearranged to given expression. |
| Answer | Marks | Guidance |
|---|---|---|
| \(CD = b\cos x\) | B1 | Correct \(CD\) in terms of \(b\) and \(x\). |
| Area \(BCD = \frac{1}{2}BC \cdot CD\sin y = \frac{1}{2}a(b\cos x)\sin y = \frac{1}{2}ab\cos x\sin y\) | B1 | Correct area of triangle \(BCD\). B0 B1 if correct area stated with no justification. |
| Area \(ABC = \frac{1}{2}AC \cdot BC\sin(x+y) = \frac{1}{2}ab\sin(x+y)\) | B1 | Correct area of triangle \(ABC\). |
| \(\frac{1}{2}ab\sin(x+y) = \frac{1}{2}ab\sin x\cos y + \frac{1}{2}ab\cos x\sin y\) | B1 | Equate area of \(ABC\) to the sum of the areas of the two small triangles and complete proof convincingly. Allow alternative proofs e.g. using lengths. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin 30\cos\alpha + \cos 30\sin\alpha = \cos 45\cos\alpha + \sin 45\sin\alpha\) | B1 | Correct use of compound angle formulae. Could be implied if exact values used immediately — allow BOD for RHS. |
| \(\frac{1}{2}\cos\alpha + \frac{1}{2}\sqrt{3}\sin\alpha = \frac{1}{2}\sqrt{2}\cos\alpha + \frac{1}{2}\sqrt{2}\sin\alpha\) | M1 | Use exact trig values. Must see all 4 values; in either equation or two expressions. |
| \((\sqrt{3}-\sqrt{2})\sin\alpha = (\sqrt{2}-1)\cos\alpha\) | M1 | Gather like terms and attempt \(\tan\alpha\). May still have fractions in the fraction. \(\tan\alpha\) does not yet need to be the subject. |
| \(\frac{\sin\alpha}{\cos\alpha} = \tan\alpha = \dfrac{\sqrt{2}-1}{\sqrt{3}-\sqrt{2}}\) | ||
| \(= \dfrac{(\sqrt{2}-1)(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})} = \dfrac{\sqrt{6}+2-\sqrt{2}-\sqrt{3}}{3-2}\) | M1 | Attempt to rationalise their denominator. Clear intention to multiply throughout by the conjugate. |
| \(\tan\alpha = 2 + \sqrt{6} - \sqrt{3} - \sqrt{2}\) A.G. | A1 | Obtain given answer. With full detail, including (at least) \(3-2\) in denominator. |
# Question 10:
## Part (a)(i)
$\cos y = \frac{CD}{a}$, hence $CD = a\cos y$ | B1 | Justification for $CD$. Need to see either $\cos y = \frac{CD}{a}$ or $\text{adj} = \text{hyp} \times \cos\theta$ before given answer.
## Part (a)(ii)
Area $= \frac{1}{2}AC \cdot CD\sin x = \frac{1}{2}b(a\cos y)\sin x = \frac{1}{2}ab\sin x\cos y$ **A.G.** | B1 | Use area of triangle to show given answer. Could quote general expression for area and then show clear substitution. Condone not being rearranged to given expression.
## Part (a)(iii)
$CD = b\cos x$ | B1 | Correct $CD$ in terms of $b$ and $x$.
Area $BCD = \frac{1}{2}BC \cdot CD\sin y = \frac{1}{2}a(b\cos x)\sin y = \frac{1}{2}ab\cos x\sin y$ | B1 | Correct area of triangle $BCD$. B0 B1 if correct area stated with no justification.
Area $ABC = \frac{1}{2}AC \cdot BC\sin(x+y) = \frac{1}{2}ab\sin(x+y)$ | B1 | Correct area of triangle $ABC$.
$\frac{1}{2}ab\sin(x+y) = \frac{1}{2}ab\sin x\cos y + \frac{1}{2}ab\cos x\sin y$ | B1 | Equate area of $ABC$ to the sum of the areas of the two small triangles and complete proof convincingly. Allow alternative proofs e.g. using lengths.
$\sin(x+y) = \sin x\cos y + \cos x\sin y$
## Part (b)
$\sin 30\cos\alpha + \cos 30\sin\alpha = \cos 45\cos\alpha + \sin 45\sin\alpha$ | B1 | Correct use of compound angle formulae. Could be implied if exact values used immediately — allow BOD for RHS.
$\frac{1}{2}\cos\alpha + \frac{1}{2}\sqrt{3}\sin\alpha = \frac{1}{2}\sqrt{2}\cos\alpha + \frac{1}{2}\sqrt{2}\sin\alpha$ | M1 | Use exact trig values. Must see all 4 values; in either equation or two expressions.
$(\sqrt{3}-\sqrt{2})\sin\alpha = (\sqrt{2}-1)\cos\alpha$ | M1 | Gather like terms and attempt $\tan\alpha$. May still have fractions in the fraction. $\tan\alpha$ does not yet need to be the subject.
$\frac{\sin\alpha}{\cos\alpha} = \tan\alpha = \dfrac{\sqrt{2}-1}{\sqrt{3}-\sqrt{2}}$ | |
$= \dfrac{(\sqrt{2}-1)(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})} = \dfrac{\sqrt{6}+2-\sqrt{2}-\sqrt{3}}{3-2}$ | M1 | Attempt to rationalise their denominator. Clear intention to multiply throughout by the conjugate.
$\tan\alpha = 2 + \sqrt{6} - \sqrt{3} - \sqrt{2}$ **A.G.** | A1 | Obtain given answer. With full detail, including (at least) $3-2$ in denominator.10
\begin{enumerate}[label=(\alph*)]
\item \\
\includegraphics[max width=\textwidth, alt={}, center]{6b766f5c-8533-4e0c-bb10-0d9949dc777b-7_599_780_267_328}
The diagram shows triangle $A B C$. The perpendicular from $C$ to $A B$ meets $A B$ at $D$. Angle $A C D = x$, angle $D C B = y$, length $B C = a$ and length $A C = b$.
\begin{enumerate}[label=(\roman*)]
\item Explain why the length of $C D$ can be written as $a \cos y$.
\item Show that the area of the triangle $A D C$ is given by $\frac { 1 } { 2 } a b \sin x \cos y$.
\item Hence, or otherwise, show that $\sin ( x + y ) = \sin x \cos y + \cos x \sin y$.
\end{enumerate}\item Given that $\sin \left( 30 ^ { \circ } + \alpha \right) = \cos \left( 45 ^ { \circ } - \alpha \right)$, show that $\tan \alpha = 2 + \sqrt { 6 } - \sqrt { 3 } - \sqrt { 2 }$.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2021 Q10 [11]}}