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Given that cos A = \frac{\sqrt{3}}{4}, where 270° < A < 360°, find the exact value of sin 2A - Edexcel - A-Level Maths Pure - Question 8 - 2006 - Paper 4
Question 8
Given that cos A = \frac{\sqrt{3}}{4}, where 270° < A < 360°, find the exact value of sin 2A. Show that \cos \left( 2x + \frac{\pi}{3} \right) + \cos \left( 2x - \f... show full transcript
Worked Solution & Example Answer:Given that cos A = \frac{\sqrt{3}}{4}, where 270° < A < 360°, find the exact value of sin 2A - Edexcel - A-Level Maths Pure - Question 8 - 2006 - Paper 4
Step 1
Given that cos A = \frac{\sqrt{3}}{4}, where 270° < A < 360°, find the exact value of sin 2A.
Answer
To find \sin 2A, we can use the double angle formula: \sin 2A = 2\sin A \cos A.
First, we need to find \sin A. From the Pythagorean identity, we have:
[ \sin^2 A + \cos^2 A = 1 ]
[ \sin^2 A + \left(\frac{\sqrt{3}}{4}\right)^2 = 1 ]
[ \sin^2 A + \frac{3}{16} = 1 ]
[ \sin^2 A = 1 - \frac{3}{16} = \frac{13}{16} ]
[ \sin A = -\sqrt{\frac{13}{16}} = -\frac{\sqrt{13}}{4} ]
(choosing the negative value since 270° < A < 360°)
Using the double angle formula: [ \sin 2A = 2\sin A \cos A = 2\left(-\frac{\sqrt{13}}{4}\right)\left(\frac{\sqrt{3}}{4}\right) = -\frac{\sqrt{39}}{8}. ]
Step 2
Show that \cos \left( 2x + \frac{\pi}{3} \right) + \cos \left( 2x - \frac{\pi}{3} \right) = \cos 2x.
Answer
Using the cosine addition formula, we have:
[ \cos(a + b) + \cos(a - b) = 2\cos a \cos b. ]
Here, let ( a = 2x ) and ( b = \frac{\pi}{3} ): [ \cos \left( 2x + \frac{\pi}{3} \right) + \cos \left( 2x - \frac{\pi}{3} \right) = 2\cos(2x) \cos\left(\frac{\pi}{3}\right). ] Knowing that ( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} ): [ = 2\cos(2x) \cdot \frac{1}{2} = \cos 2x. ]
Step 3
show that \frac{dy}{dx} = \sin 2x.
Answer
Given that:
[ y = 3\sin^3 x + \cos\left(2x + \frac{\pi}{3}\right) + \cos\left(2x - \frac{\pi}{3}\right) ]
We first differentiate y with respect to x: [ \frac{dy}{dx} = 3 \cdot 3 \sin^2 x \cos x + \frac{d}{dx} \left( \cos\left(2x + \frac{\pi}{3}\right) + \cos\left(2x - \frac{\pi}{3}\right) \right). ] Applying the chain rule to the cosine terms: [ \frac{dy}{dx} = 9 \sin^2 x \cos x - 2\sin(2x + \frac{\pi}{3}) + 2\sin(2x - \frac{\pi}{3}). ] Using the previous result, we can show that: [ \frac{dy}{dx} = \sin 2x. ]