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Given that cos A = \frac{\sqrt{7}}{4}, where 270° < A < 360°, find the exact value of sin 2A - Edexcel - A-Level Maths Pure - Question 2 - 2006 - Paper 5
Question 2
Given that cos A = \frac{\sqrt{7}}{4}, where 270° < A < 360°, find the exact value of sin 2A. Show that \cos \left( 2x + \frac{\pi}{3} \right) + \cos \left( 2x - ... show full transcript
Worked Solution & Example Answer:Given that cos A = \frac{\sqrt{7}}{4}, where 270° < A < 360°, find the exact value of sin 2A - Edexcel - A-Level Maths Pure - Question 2 - 2006 - Paper 5
Step 1
Given that cos A = \frac{\sqrt{7}}{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 ]
Since we know \cos A = \frac{\sqrt{7}}{4}$, we need to find \sin A.
Using the identity ( \sin^2 A + \cos^2 A = 1 ):
[ \sin^2 A = 1 - \cos^2 A = 1 - \left(\frac{\sqrt{7}}{4}\right)^2 = 1 - \frac{7}{16} = \frac{9}{16} ]
Thus, ( \sin A = \frac{3}{4} ) (since A is in the fourth quadrant, ( \sin A ) is negative, so we take ( \sin A = -\frac{3}{4} )).
Therefore,
[ \sin 2A = 2 \cdot \left(-\frac{3}{4}\right) \cdot \frac{\sqrt{7}}{4} = -\frac{6\sqrt{7}}{16} = -\frac{3\sqrt{7}}{8}. ]
Step 2
Show that \cos \left( 2x + \frac{\pi}{3} \right) + \cos \left( 2x - \frac{\pi}{3} \right) = \cos 2x.
Answer
We can use the cosine sum-to-product identities:
[ \cos a + \cos b = 2 \cos \left( \frac{a + b}{2} \right) \cos \left( \frac{a - b}{2} \right). ]
Letting ( a = 2x + \frac{\pi}{3} ) and ( b = 2x - \frac{\pi}{3} ):
[ \frac{a + b}{2} = \frac{(2x + \frac{\pi}{3}) + (2x - \frac{\pi}{3})}{2} = \frac{4x}{2} = 2x, ]
[ \frac{a - b}{2} = \frac{(2x + \frac{\pi}{3}) - (2x - \frac{\pi}{3})}{2} = \frac{\frac{2\pi}{3}}{2} = \frac{\pi}{3}. ]
Thus,
[ \cos \left( 2x + \frac{\pi}{3} \right) + \cos \left( 2x - \frac{\pi}{3} \right) = 2 \cos(2x) \cos \left( \frac{\pi}{3} \right) = 2 \cos(2x) \cdot \frac{1}{2} = \cos 2x. ]
Step 3
show that \frac{dy}{dx} = \sin 2x.
Answer
Given [ y = 3 \sin^2 x + \cos \left( 2x + \frac{\pi}{3} \right) + \cos \left( 2x - \frac{\pi}{3} \right), ] we differentiate term by term.
Using the derivative of ( \sin^2 x ) which is ( 2 \sin x \cos x ) and the derivative of ( \cos(2x) ) which is ( -2 \sin(2x) ):
[ \frac{dy}{dx} = 3 \cdot 2 \sin x \cos x - 2 \sin(2x) ]
Recognizing that ( 2 \sin x \cos x = \sin(2x) ), we simplify the expression:
[ \frac{dy}{dx} = 3 \sin(2x) - 2 \sin(2x) = \sin(2x). ]