| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2020 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Express and solve equation |
| Difficulty | Standard +0.3 This is a standard harmonic form question requiring routine application of the R cos(θ - α) formula with R = √(6+9) = √15 and tan α = 3/√6, followed by solving a straightforward equation with a substitution u = x/3. The only minor complication is handling the angle transformation, but this is a well-practiced technique at this level. |
| Spec | 1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State \(R = \sqrt{15}\) | B1 | |
| Use trig formulae to find \(\alpha\) | M1 | \(\frac{\sin\alpha}{\cos\alpha} = \frac{3}{\sqrt{6}}\) with no error seen or \(\tan\alpha = \frac{3}{\sqrt{6}}\) quoted then allow M1 |
| Obtain \(\alpha = 50.77\) | A1 | Must be 2 d.p. If radians 0.89: A0 MR |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Evaluate \(\beta = \cos^{-1}\frac{2.5}{\sqrt{15}}\) (49.797° to 4 d.p.) | B1 FT | The FT is on incorrect \(R\). \(\frac{x}{3} = \beta - \alpha\) \([-2.9°\) and \(-301.7°]\) |
| Use correct method to find a value of \(\frac{x}{3}\) in the interval | M1 | Needs to use \(\frac{x}{3}\) |
| Obtain answer rounding to \(x = 301.6°\) to \(301.8°\) | A1 | |
| Obtain second answer rounding to \(x = 2.9(0)°\) to \(2.9(2)°\) and no others in the interval | A1 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State $R = \sqrt{15}$ | B1 | |
| Use trig formulae to find $\alpha$ | M1 | $\frac{\sin\alpha}{\cos\alpha} = \frac{3}{\sqrt{6}}$ with no error seen or $\tan\alpha = \frac{3}{\sqrt{6}}$ quoted then allow M1 |
| Obtain $\alpha = 50.77$ | A1 | Must be 2 d.p. If radians 0.89: A0 MR |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Evaluate $\beta = \cos^{-1}\frac{2.5}{\sqrt{15}}$ (49.797° to 4 d.p.) | B1 FT | The FT is on incorrect $R$. $\frac{x}{3} = \beta - \alpha$ $[-2.9°$ and $-301.7°]$ |
| Use correct method to find a value of $\frac{x}{3}$ in the interval | M1 | Needs to use $\frac{x}{3}$ |
| Obtain answer rounding to $x = 301.6°$ to $301.8°$ | A1 | |
| Obtain second answer rounding to $x = 2.9(0)°$ to $2.9(2)°$ and no others in the interval | A1 | |6
\begin{enumerate}[label=(\alph*)]
\item Express $\sqrt { 6 } \cos \theta + 3 \sin \theta$ in the form $R \cos ( \theta - \alpha )$, where $R > 0$ and $0 ^ { \circ } < \alpha < 90 ^ { \circ }$. State the exact value of $R$ and give $\alpha$ correct to 2 decimal places.
\item Hence solve the equation $\sqrt { 6 } \cos \frac { 1 } { 3 } x + 3 \sin \frac { 1 } { 3 } x = 2.5$, for $0 ^ { \circ } < x < 360 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2020 Q6 [7]}}