5 Given that \(\sin ( \theta + \alpha ) = 2 \sin \theta\), show that \(\tan \theta = \frac { \sin \alpha } { 2 - \cos \alpha }\).
Hence solve the equation \(\sin \left( \theta + 40 ^ { \circ } \right) = 2 \sin \theta\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).
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Question 5:
Answer Marks
Guidance
Answer Marks
Guidance
\(\sin(\theta+\alpha) = 2\sin\theta\) \(\Rightarrow \sin\theta\cos\alpha + \cos\theta\sin\alpha = 2\sin\theta\) M1
Using correct compound angle formula in a valid equation
\(\Rightarrow \tan\theta\cos\alpha + \sin\alpha = 2\tan\theta\) M1
Dividing by \(\cos\theta\)
\(\Rightarrow \sin\alpha = 2\tan\theta - \tan\theta\cos\alpha = \tan\theta(2-\cos\alpha)\) M1
Collecting terms in \(\tan\theta\) or \(\sin\theta\) or dividing by \(\tan\theta\) oe
\(\Rightarrow \tan\theta = \frac{\sin\alpha}{2-\cos\alpha}\) E1
www (can be all achieved for the method in reverse)
\(\sin(\theta+40°) = 2\sin\theta \Rightarrow \tan\theta = \frac{\sin 40}{2-\cos 40} = 0.5209\) M1
\(\tan\theta = \frac{\sin 40}{2-\cos 40}\)
\(\Rightarrow \theta = 27.5°, 207.5°\) A1 A1
\(-1\) if given in radians. \(-1\) extra solutions in range.
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## Question 5:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sin(\theta+\alpha) = 2\sin\theta$ | | |
| $\Rightarrow \sin\theta\cos\alpha + \cos\theta\sin\alpha = 2\sin\theta$ | M1 | Using correct compound angle formula in a valid equation |
| $\Rightarrow \tan\theta\cos\alpha + \sin\alpha = 2\tan\theta$ | M1 | Dividing by $\cos\theta$ |
| $\Rightarrow \sin\alpha = 2\tan\theta - \tan\theta\cos\alpha = \tan\theta(2-\cos\alpha)$ | M1 | Collecting terms in $\tan\theta$ or $\sin\theta$ or dividing by $\tan\theta$ oe |
| $\Rightarrow \tan\theta = \frac{\sin\alpha}{2-\cos\alpha}$ | E1 | www (can be all achieved for the method in reverse) |
| $\sin(\theta+40°) = 2\sin\theta \Rightarrow \tan\theta = \frac{\sin 40}{2-\cos 40} = 0.5209$ | M1 | $\tan\theta = \frac{\sin 40}{2-\cos 40}$ |
| $\Rightarrow \theta = 27.5°, 207.5°$ | A1 A1 | $-1$ if given in radians. $-1$ extra solutions in range. |
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5 Given that $\sin ( \theta + \alpha ) = 2 \sin \theta$, show that $\tan \theta = \frac { \sin \alpha } { 2 - \cos \alpha }$.
Hence solve the equation $\sin \left( \theta + 40 ^ { \circ } \right) = 2 \sin \theta$, for $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.
\hfill \mbox{\textit{OCR MEI C4 Q5 [7]}}