Edexcel C2 2010 June — Question 5 6 marks

Exam BoardEdexcel
ModuleC2 (Core Mathematics 2)
Year2010
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTrigonometric equations in context
TypeConvert sin/cos ratio to tan
DifficultyModerate -0.3 Part (a) is a straightforward application of the definition tan θ = sin θ/cos θ requiring simple algebraic manipulation. Part (b) extends this to a double angle equation, requiring students to recognize the connection to part (a) and account for the period of 2x, but remains a standard C2 exercise with clear scaffolding from part (a).
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals

5. (a) Given that \(5 \sin \theta = 2 \cos \theta\), find the value of \(\tan \theta\).
(b) Solve, for \(0 \leqslant x < 360 ^ { \circ }\), $$5 \sin 2 x = 2 \cos 2 x$$ giving your answers to 1 decimal place.

Question 5:
Part (a):
AnswerMarks Guidance
AnswerMarks Guidance
\(\tan\theta = \frac{2}{5}\) (or 0.4)B1 Requires correct value with no incorrect working seen. i.s.w. if a value of \(\theta\) is subsequently found
Part (b):
AnswerMarks Guidance
AnswerMarks Guidance
awrt \(21.8\) \((\alpha)\)B1 Also allow awrt 68.2 from \(\tan\theta=\frac{5}{2}\) in (a), but no other. Must be seen in part (b)
\(180+\alpha\) (= 201.8) or \(90+(\alpha/2)\)M1 \(\alpha\) found from \(\tan 2x=\ldots\) or \(\tan x=\ldots\) or \(\sin 2x=\pm\ldots\) or \(\cos 2x=\pm\ldots\)
\(360+\alpha\) (= 381.8) or \(180+(\alpha/2)\)M1 \(\alpha\) found from \(\tan 2x=\ldots\) or \(\sin 2x=\ldots\) or \(\cos 2x=\ldots\)
Dividing at least one angle by 2M1 \(\alpha\) found from \(\tan 2x=\ldots\) or \(\sin 2x=\ldots\) or \(\cos 2x=\ldots\)
\(x = 10.9, 100.9, 190.9, 280.9\)A1 Allow awrt. Extra solutions in range loses final A mark
## Question 5:

### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\tan\theta = \frac{2}{5}$ (or 0.4) | B1 | Requires correct value with no incorrect working seen. i.s.w. if a value of $\theta$ is subsequently found |

### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| awrt $21.8$ $(\alpha)$ | B1 | Also allow awrt 68.2 from $\tan\theta=\frac{5}{2}$ in (a), but no other. Must be seen in part (b) |
| $180+\alpha$ (= 201.8) or $90+(\alpha/2)$ | M1 | $\alpha$ found from $\tan 2x=\ldots$ or $\tan x=\ldots$ or $\sin 2x=\pm\ldots$ or $\cos 2x=\pm\ldots$ |
| $360+\alpha$ (= 381.8) or $180+(\alpha/2)$ | M1 | $\alpha$ found from $\tan 2x=\ldots$ or $\sin 2x=\ldots$ or $\cos 2x=\ldots$ |
| Dividing at least one angle by 2 | M1 | $\alpha$ found from $\tan 2x=\ldots$ or $\sin 2x=\ldots$ or $\cos 2x=\ldots$ |
| $x = 10.9, 100.9, 190.9, 280.9$ | A1 | Allow awrt. Extra solutions in range loses final A mark |

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5. (a) Given that $5 \sin \theta = 2 \cos \theta$, find the value of $\tan \theta$.\\
(b) Solve, for $0 \leqslant x < 360 ^ { \circ }$,

$$5 \sin 2 x = 2 \cos 2 x$$

giving your answers to 1 decimal place.\\

\hfill \mbox{\textit{Edexcel C2 2010 Q5 [6]}}