When \( \theta \) is small, find an approximation for \( \cos 30^{\circ} + \theta \sin 2\theta \), giving your answer in the form \( a + b\theta^2 \).
- AQA - A-Level Maths Pure - Question 3 - 2017 - Paper 1
Question 3
When \( \theta \) is small, find an approximation for \( \cos 30^{\circ} + \theta \sin 2\theta \), giving your answer in the form \( a + b\theta^2 \).
Worked Solution & Example Answer:When \( \theta \) is small, find an approximation for \( \cos 30^{\circ} + \theta \sin 2\theta \), giving your answer in the form \( a + b\theta^2 \).
- AQA - A-Level Maths Pure - Question 3 - 2017 - Paper 1
Step 1
Use approximations for cosine and sine
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Answer
For small angles, we use the approximations:
( \cos x \approx 1 - \frac{x^2}{2} )
( \sin x \approx x ).
In this case, we apply this to ( \cos 30^{\circ} ) and ( \sin 2\theta ):
( \cos 30^{\circ} = \frac{\sqrt{3}}{2} ) which can be approximated directly since it is a constant: ( \approx \frac{\sqrt{3}}{2} )
( \sin 2\theta \approx 2\theta ).
Step 2
Substitutes 2\( \theta \) and 30\( \theta \) into ‘their’ expression
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Answer
Substituting these approximations into the expression:
[ \cos 30^{\circ} + \theta \sin 2\theta \approx \frac{\sqrt{3}}{2} + \theta(2\theta) = \frac{\sqrt{3}}{2} + 2\theta^2.]
Step 3
Obtain correct answer
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Answer
Since we want the answer in the form ( a + b\theta^2 ), we compare:
( a = \frac{\sqrt{3}}{2} ) and ( b = 2 ).
Thus, the final approximation is:
[ \frac{\sqrt{3}}{2} + 2\theta^2. ]
