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. (4) (b) (i) Prove that $$ \frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x} = 2 \sec x, \quad x \neq (2n + 1)\frac{\pi}{2}. $$ (4) (ii) Hence find, for $0 < x < 2\pi$, all the solutions of $$ \frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x} = 4. $$ (3)

A-Level Edexcel Maths Pure Trigonometric Functions: 6. (a) Use the double angle form

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

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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... 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

Use the double angle formulae and the identity to obtain an expression for cos 3x

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Answer

To find an expression for $\cos 3x$ in terms of $\cos x$ only, we can use the double angle identity:

\cos (A+B) = \cos A \cos B - \sin A \sin B.

Let us first express $\cos 3x$ as:

\cos 3x = \cos (2x + x) = \cos 2x \cos x - \sin 2x \sin x.

Next, we apply the double angle formulas:

$\cos 2x = 2\cos^2 x - 1$
$\sin 2x = 2\sin x \cos x$

Substituting these into our expression gives:

\cos 3x = (2\cos^2 x - 1)\cos x - (2\sin x \cos x)\sin x.

Now, notice that:

$\sin^2 x = 1 - \cos^2 x$

Thus, we have:

\cos 3x = 2\cos^3 x - \cos x - 2\sin^2 x \cos x = 2\cos^3 x - \cos x - 2(1 - \cos^2 x)\cos x.

This simplifies to:

\cos 3x = 4\cos^3 x - 3\cos x.

Step 2

Prove that \frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x} = 2 \sec x

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Answer

Let's simplify the left side:

\frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x}.

Finding a common denominator gives:

= \frac{\cos^2 x + (1 + \sin x)^2}{(1 + \sin x)\cos x}.

Now expand the numerator:

= \frac{\cos^2 x + 1 + 2\sin x + \sin^2 x}{(1 + \sin x)\cos x}.

Using the Pythagorean identity $\cos^2 x + \sin^2 x = 1$ gives us:

= \frac{1 + 1 + 2\sin x}{(1 + \sin x)\cos x} = \frac{2(1 + \sin x)}{(1 + \sin x)\cos x}.

Now canceling $1 + \sin x$ leads us to:

= \frac{2}{\cos x} = 2 \sec x,

which proves the statement.

Step 3

Hence find, for 0 < x < 2π, all the solutions of \frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x} = 4

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Answer

From the previous result, we know 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 \implies \sec x = 2.

This implies:

\cos x = \frac{1}{2}.

The solutions for $\cos x = \frac{1}{2}$ within the interval $0 < x < 2\pi$ are:

\begin{align*} x &= \frac{\pi}{3}, \\ x &= \frac{5\pi}{3}. \end{align*}$$

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

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

Use the double angle formulae and the identity to obtain an expression for cos 3x

Prove that \frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x} = 2 \sec x

Hence find, for 0 < x < 2π, all the solutions of \frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x} = 4

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