6. (a) Use the double angle formulae and the identity $$ ext{cos}(A+B) = ext{cos} A ext{cos} B - ext{sin} A ext{sin} B $$ to obtain an expression for cos 3x in terms of powers of cos x only. (b) (i) Prove that $$ rac{ ext{cos} x}{1+ ext{sin} x} + rac{1+ ext{sin} x}{ ext{cos} x} = 2 ext{sec} x, ext{ } x eq (2n+1)rac{ ext{π}}{2}. $$ (ii) Hence find, for 0 < x < 2π, all the solutions of $$ rac{ ext{cos} x}{1+ ext{sin} x} + rac{1+ ext{sin} x}{ ext{cos} x} = 4.

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

6. (a) Use the double angle formulae and the identity $$ ext{cos}(A+B) = ext{cos} A ext{cos} B - ext{sin} A ext{sin} B $$ to obtain an expression for cos 3x in terms of powers of cos x only - Edexcel - A-Level Maths Pure - Question 6 - 2008 - Paper 6

Question 6

6.-(a)-Use-the-double-angle-formulae-and-the-identity--$$--ext{cos}(A+B)-=--ext{cos}-A--ext{cos}-B----ext{sin}-A--ext{sin}-B-$$--to-obtain-an-expression-for-cos-3x-in-terms-of-powers-of-cos-x-only-Edexcel-A-Level Maths Pure-Question 6-2008-Paper 6.png

6. (a) Use the double angle formulae and the identity $$ ext{cos}(A+B) = ext{cos} A ext{cos} B - ext{sin} A ext{sin} B $$ to obtain an expression for cos 3x i... show full transcript

Worked Solution & Example Answer:6. (a) Use the double angle formulae and the identity $$ ext{cos}(A+B) = ext{cos} A ext{cos} B - ext{sin} A ext{sin} B $$ to obtain an expression for cos 3x in terms of powers of cos x only - Edexcel - A-Level Maths Pure - Question 6 - 2008 - Paper 6

Step 1

Use the double angle formulae and the identity

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Answer

Starting with the double angle formula for cosine:

ext{cos}(2x) = 2 ext{cos}^2 x - 1,

we can express cos(3x) as:

ext{cos}(3x) = ext{cos}(2x + x) = ext{cos}(2x) ext{cos}(x) - ext{sin}(2x) ext{sin}(x).

Substituting cos(2x) and sin(2x):

= (2 ext{cos}^2 x - 1) ext{cos} x - (2 ext{sin} x ext{cos} x) ext{sin} x.

Using the identity $ext{sin}^2 x = 1 - ext{cos}^2 x$ gives:

= (2 ext{cos}^2 x - 1) ext{cos} x - (2 ext{sin}^2 x) ext{cos} x = 2 ext{cos}^3 x - ext{cos} x - 2(1 - ext{cos}^2 x) ext{cos} x.

Simplifying, we arrive at:

$$ ext{cos} 3x = 4 ext{cos}^3 x - 3 ext{cos} x.$

Step 2

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

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Answer

Starting with the left-hand side:

rac{ ext{cos} x}{1 + ext{sin} x} + rac{1 + ext{sin} x}{ ext{cos} x} = rac{ ext{cos}^2 x + (1 + ext{sin} x)(1 + ext{sin} x)}{(1 + ext{sin} x) ext{cos} x}.

Now simplifying the numerator:

Thus, we have:

This proves the statement.

Step 3

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

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Answer

From the proven expression:

$2 ext{sec} x = 4,$ we can deduce:

$ext{sec} x = 2.$

This implies:

$ext{cos} x = rac{1}{2}.$

The general solutions for this equation are:

$x = rac{ ext{π}}{3} + 2n ext{π} ext{ and } x = rac{5 ext{π}}{3} + 2n ext{π}$ for integer n. We restrict n to the range such that (0 < x < 2π).

Thus, the specific solutions for 0 < x < 2π are:

$$x = rac{ ext{π}}{3} ext{ and } x = rac{5 ext{π}}{3}.$

6. (a) Use the double angle formulae and the identity $$ ext{cos}(A+B) = ext{cos} A ext{cos} B - ext{sin} A ext{sin} B $$ to obtain an expression for cos 3x in terms of powers of cos x only - Edexcel - A-Level Maths Pure - Question 6 - 2008 - Paper 6

Worked Solution & Example Answer:6. (a) Use the double angle formulae and the identity $$ ext{cos}(A+B) = ext{cos} A ext{cos} B - ext{sin} A ext{sin} B $$ to obtain an expression for cos 3x in terms of powers of cos x only - Edexcel - A-Level Maths Pure - Question 6 - 2008 - Paper 6

Use the double angle formulae and the identity

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

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

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