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6. (a) Use the double angle formulae and the identity \( \cos(A+B) = \cos A \cos B - \sin A \sin B \) to obtain an expression for \( \cos 3x \) in terms of powers of \( \cos x \) only - Edexcel - A-Level Maths Pure - Question 8 - 2008 - Paper 6
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6. (a) Use the double angle formulae and the identity \( \cos(A+B) = \cos A \cos B - \sin A \sin B \) to obtain an expression for \( \cos 3x \) in terms of powers o... show full transcript
Worked Solution & Example Answer:6. (a) Use the double angle formulae and the identity \( \cos(A+B) = \cos A \cos B - \sin A \sin B \) to obtain an expression for \( \cos 3x \) in terms of powers of \( \cos x \) only - Edexcel - A-Level Maths Pure - Question 8 - 2008 - Paper 6
Step 1
Use the double angle formulae and the identity \( \cos(A+B) \)
Answer
To find ( \cos 3x ), we start with the double angle formula.
- Express ( \cos 3x ) using the identity: [ \cos 3x = \cos(2x + x) = \cos 2x \cos x - \sin 2x \sin x ]
- Substitute the double angle formulas for ( \cos 2x ) and ( \sin 2x ): [ \cos 2x = 2\cos^2 x - 1 \quad \text{and} \quad \sin 2x = 2\sin x \cos x ]
- Substitute these into the expression: [ \cos 3x = (2\cos^2 x - 1)\cos x - 2\sin x \cos x \sin x ]
- Rearranging gives: [ \cos 3x = 2\cos^3 x - \cos x - 2\sin^2 x \cos x ]
- Using the identity ( \sin^2 x = 1 - \cos^2 x ): [ \cos 3x = 2\cos^3 x - \cos x - 2(1 - \cos^2 x)\cos x ]
- Simplifying the expression leads to: [ \cos 3x = 4\cos^3 x - 3\cos x ]
Step 2
(i) Prove that \( \frac{\cos x}{1+\sin x} + \frac{1+\sin x}{\cos x} = 2 \sec x \)
Answer
To prove the equation, we will find a common denominator and simplify:
- Combine the fractions: [ \frac{\cos x (\cos x)}{(1+\sin x)\cos x} + \frac{(1+\sin x)(1+\sin x)}{(1+\sin x)\cos x} = \frac{\cos^2 x + (1+\sin x)^2}{(1+\sin x)\cos x} ]
- Now, simplify the numerator: [ \cos^2 x + 1 + 2\sin x + \sin^2 x = 1 + 1 + 2\sin x = 2 + 2\sin x ]
- Thus, we have: [ \frac{2 + 2\sin x}{(1+\sin x)\cos x} = \frac{2(1 + \sin x)}{(1 + \sin x)\cos x} ]
- Cancel out the common terms: [ = \frac{2}{\cos x} = 2 \sec x ]
Step 3
(ii) Hence find, for \( 0 < x < 2\pi \), all the solutions of \( \frac{\cos x}{1+\sin x} + \frac{1+\sin x}{\cos x} = 4. \)
Answer
From part (i), we have established that: [ \frac{\cos x}{1+\sin x} + \frac{1+\sin x}{\cos x} = 2 \sec x ] Setting this equal to 4 gives: [ 2 \sec x = 4 ] [ \sec x = 2 ] 5. Thus, ( \cos x = \frac{1}{2} ) 6. The possible solutions in the range ( 0 < x < 2\pi ) are: [ x = \frac{\pi}{3}, \quad x = \frac{5\pi}{3} ]