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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 7 - 2008 - Paper 6
Question 7
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. ... 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 7 - 2008 - Paper 6
Step 1
(a) Obtain an expression for cos 3x in terms of powers of cos x only.
Answer
To find an expression for ( \cos 3x ) using the double angle formulae, we start by using the identity:
[ \cos(3x) = \cos(2x + x) = \cos 2x \cos x - \sin 2x \sin x. ]
From the double angle formula:
- ( \cos 2x = 2 \cos^2 x - 1 )
- ( \sin 2x = 2 \sin x \cos x )
Substituting these into the expression gives:
[ \cos(3x) = (2 \cos^2 x - 1) \cos x - (2 \sin x \cos x) \sin x. ]
This simplifies to:
[ \cos(3x) = 2 \cos^3 x - \cos x - 2 \sin^2 x \cos x. ]
Using the identity ( \sin^2 x = 1 - \cos^2 x ), we can substitute for ( \sin^2 x ):
[ \cos(3x) = 2 \cos^3 x - \cos x - 2(1 - \cos^2 x)\cos x. ]
Simplifying further, we obtain:
[ \cos(3x) = 4 \cos^3 x - 3 \cos x. ]
Step 2
(b) (i) Prove that \(\frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x} = 2 \sec x\).
Answer
To prove the identity, consider the expression:
[ LHS = \frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x}. ]
Finding a common denominator, we get:
[ LHS = \frac{\cos^2 x + (1 + \sin x)^2}{(1 + \sin x) \cos x} = \frac{\cos^2 x + 1 + 2\sin x + \sin^2 x}{(1 + \sin x) \cos x}. ]
Using the Pythagorean identity ( \cos^2 x + \sin^2 x = 1 ), we simplify:
[ LHS = \frac{1 + 1 + 2\sin x}{(1 + \sin x) \cos x} = \frac{2(1 + \sin x)}{(1 + \sin x) \cos x} = \frac{2}{\cos x} = 2 \sec x. ]
Thus, we've proven the identity.
Step 3
(b) (ii) Hence find, for 0 < x < 2π, all the solutions of \(\frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x} = 4.\)
Answer
We start from the proven identity:
[ \frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x} = 2 \sec x. ]
Now set this equal to 4:
[ 2 \sec x = 4 \implies \sec x = 2. ]
This leads to:
[ \cos x = \frac{1}{2}. ]
The solutions for ( \cos x = \frac{1}{2} ) over the interval (0 < x < 2\pi) occurs at:
[ x = \frac{\pi}{3}, \quad x = \frac{5\pi}{3}. ]
Thus, the solutions are ( x = \frac{\pi}{3}, \frac{5\pi}{3} ).