| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Graphs & Exact Values |
| Type | Find exact trig values from given ratio |
| Difficulty | Moderate -0.3 This question tests standard trig identities and exact values with straightforward application. Part (a) involves recognizing sin(a)=0.5 gives a=30°, then using the supplementary angle identity. Part (b) requires the sine difference identity and Pythagorean identity to find cosine from a given sine value. While multi-part, each step follows directly from basic A-level trig knowledge without requiring problem-solving insight or complex manipulation. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| \(-180 - 30 = -210\) | M1 | States or uses \(\sin 30 = 0.5\); PI by sight of \(\pm 150\) or \(\pm 30\) or \(-330\); maybe seen on diagram |
| \(-210\) | A1 | Obtains \(-210\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.5\) | B1 | Obtains \(0.5\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin(b-180) = -\sin b = \frac{3}{7}\) | M1 | Uses correct approach to find \(\sin(b-180)\); might see \(\sin(205.37...\pm 180)\) or \(\sin(\pm 180 - 25.376...)\) or correct use of compound angle formula |
| \(\sin(b-180) = \frac{3}{7}\) | R1 | Deduces \(\sin(b-180) = \frac{3}{7}\); CAO |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos^2 b + \left(-\frac{3}{7}\right)^2 = 1\); \(\cos^2 b = \frac{40}{49}\) | M1 | Uses \(\cos^2 x + \sin^2 x = 1\) or draws right angled triangle with 3 and 7 on opposite and hypotenuse sides; PI by \(\cos b = -\frac{2\sqrt{10}}{7}\) OE exact form |
| \(\cos^2 b = \frac{40}{49}\) | A1 | Condone \(b\) replaced by different variable or obtains ratio for cosine of correct exact magnitude |
| \(\cos b = -\frac{2\sqrt{10}}{7}\) | R1 | Deduces \(\cos b = -\frac{2\sqrt{10}}{7}\); OE exact form; CAO |
## Question 10:
### 10(a)(i):
$-180 - 30 = -210$ | M1 | States or uses $\sin 30 = 0.5$; PI by sight of $\pm 150$ or $\pm 30$ or $-330$; maybe seen on diagram
$-210$ | A1 | Obtains $-210$
### 10(a)(ii):
$0.5$ | B1 | Obtains $0.5$
### 10(b)(i):
$\sin(b-180) = -\sin b = \frac{3}{7}$ | M1 | Uses correct approach to find $\sin(b-180)$; might see $\sin(205.37...\pm 180)$ or $\sin(\pm 180 - 25.376...)$ or correct use of compound angle formula
$\sin(b-180) = \frac{3}{7}$ | R1 | Deduces $\sin(b-180) = \frac{3}{7}$; CAO
### 10(b)(ii):
$\cos^2 b + \left(-\frac{3}{7}\right)^2 = 1$; $\cos^2 b = \frac{40}{49}$ | M1 | Uses $\cos^2 x + \sin^2 x = 1$ or draws right angled triangle with 3 and 7 on opposite and hypotenuse sides; PI by $\cos b = -\frac{2\sqrt{10}}{7}$ OE exact form
$\cos^2 b = \frac{40}{49}$ | A1 | Condone $b$ replaced by different variable or obtains ratio for cosine of correct exact magnitude
$\cos b = -\frac{2\sqrt{10}}{7}$ | R1 | Deduces $\cos b = -\frac{2\sqrt{10}}{7}$; OE exact form; CAO
---10
\begin{enumerate}[label=(\alph*)]
\item Point $A$ on the curve has coordinates ( $a , 0.5$ )\\
10 (a) (i) Find the value of $a$\\
10 (a) (ii) State the value of $\sin \left( 180 ^ { \circ } - a ^ { \circ } \right)$\\
10
\item Point $B$ on the curve has coordinates $\left( b , - \frac { 3 } { 7 } \right)$\\
10 (b) (i) Find the exact value of $\sin \left( b ^ { \circ } - 180 ^ { \circ } \right)$\\
10 (b) (ii) Find the exact value of $\cos b ^ { \circ }$
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2023 Q10 [8]}}