9. (a) Express 2 sin θ - 4 cos θ in the form R sin(θ - α), where R and α are constants, R > 0 and 0 < α < π/2. Give the value of α to 3 decimal places. H(θ) = 4 + 5(2 sin 3θ - 4 cos 3θ)² Find (b) (i) the maximum value of H(θ), (ii) the smallest value of θ, for 0 ≤ θ < π, at which this maximum value occurs. (c) (i) the minimum value of H(θ), (ii) the largest value of θ, for 0 ≤ θ < π, at which this minimum value occurs.

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9. (a) Express 2 sin θ - 4 cos θ in the form R sin(θ - α), where R and α are constants, R > 0 and 0 < α < π/2 - Edexcel - A-Level Maths Pure - Question 2 - 2014 - Paper 6

Question 2

9. (a) Express 2 sin θ - 4 cos θ in the form R sin(θ - α), where R and α are constants, R > 0 and 0 < α < π/2. Give the value of α to 3 decimal places. H(θ) = 4 + ... show full transcript

Worked Solution & Example Answer:9. (a) Express 2 sin θ - 4 cos θ in the form R sin(θ - α), where R and α are constants, R > 0 and 0 < α < π/2 - Edexcel - A-Level Maths Pure - Question 2 - 2014 - Paper 6

Step 1

Express 2 sin θ - 4 cos θ in the form R sin(θ - α)

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Answer

To express the equation in the desired form, we first identify constants R and α.

We start by comparing it with the formula:

$R ext{sin}( heta - eta) = R ( ext{sin}( heta) ext{cos}(eta) - ext{cos}( heta) ext{sin}(eta))$

This indicates that:

2 = R cos(α)
-4 = R sin(α).

Next, we find R using:

ightarrow R = oot{20} = rac{2 oot{5}}{1}.$$ Then, to find α, we compute: $$ an(α) = rac{-4}{2} = -2 ightarrow α = an^{-1}(-2) ightarrow α = ext{atan}(2) ≈ 1.107 \ (to \ 3 \ decimal \ places).$$ Thus, we have $R = oot{20}$ and $α ≈ 1.107$ to three decimal places.

Step 2

the maximum value of H(θ)

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Answer

To find the maximum value of H(θ), we first need to simplify the given equation:

$H(θ) = 4 + 5(2 ext{sin}(3θ) - 4 ext{cos}(3θ))^2$

The expression $(2 ext{sin}(3θ) - 4 ext{cos}(3θ))$ attains its maximum value when:

$2 ext{sin}(3θ) - 4 ext{cos}(3θ) = R ext{sin}(3θ - α'),$

where $R$ can be calculated as:

oot{(2^2) + (-4)^2} = oot{20}.$$ The maximum value is therefore: $$ ext{max}(H(θ)) = 4 + 5(R)^2 = 4 + 5 imes 20 = 104.$$

Step 3

the smallest value of θ for 0 ≤ θ < π, at which this maximum value occurs

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Answer

Using the earlier expression for $2 ext{sin}(3θ) - 4 ext{cos}(3θ)$ , we need to solve:

$2 ext{sin}(3θ) - 4 ext{cos}(3θ) = R$

This occurs when:

$an(3θ) = rac{4}{2} = 2.$

Thus:

$3θ = an^{-1}(2) + n ext{π}, ext{ where } n ext{ is an integer}.$

This gives:

$θ = rac{1}{3}( an^{-1}(2) + n ext{π}).$

For $0 ≤ θ < π$ , the smallest value occurs at $n = 0,$ hence:

$θ = rac{1}{3} an^{-1}(2) ≈ 0.89.$

Step 4

the minimum value of H(θ)

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Answer

The minimum value of H(θ) will occur when $2 ext{sin}(3θ) - 4 ext{cos}(3θ)$ is at its minimum. Since $-R$ will provide this minimum, we calculate:

$H(θ)_{ ext{min}} = 4 + 5(-R)^2 = 4 + 5(20) = 4.$

Step 5

the largest value of θ for 0 ≤ θ < π, at which this minimum value occurs

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Answer

Using the minimum condition:

$2 ext{sin}(3θ) - 4 ext{cos}(3θ) = -R,$

We have:

$an(3θ) = rac{-4}{2} = -2.$

Solving yields:

$3θ = an^{-1}(-2) + n ext{π}.$

This again leads to:

$θ = rac{1}{3}( an^{-1}(-2) + n ext{π}).$

The largest value of θ occurs when $n = 1: $$θ = rac{1}{3}( an^{-1}(-2) + π) = rac{π + 3θ}{3}.$$$

9. (a) Express 2 sin θ - 4 cos θ in the form R sin(θ - α), where R and α are constants, R > 0 and 0 < α < π/2 - Edexcel - A-Level Maths Pure - Question 2 - 2014 - Paper 6

Worked Solution & Example Answer:9. (a) Express 2 sin θ - 4 cos θ in the form R sin(θ - α), where R and α are constants, R > 0 and 0 < α < π/2 - Edexcel - A-Level Maths Pure - Question 2 - 2014 - Paper 6

Express 2 sin θ - 4 cos θ in the form R sin(θ - α)

the maximum value of H(θ)

the smallest value of θ for 0 ≤ θ < π, at which this maximum value occurs

the minimum value of H(θ)

the largest value of θ for 0 ≤ θ < π, at which this minimum value occurs

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