Using the identity cos(A + B) = cos A cos B - sin A sin B, prove that cos 2A = 1 - 2 sin² A. Show that 2 sin 2θ - 3 cos 2θ - 3 sin θ + 3 = sin θ(4 cos θ + 6 sin θ - 3). Express 4 cos θ + 6 sin θ in the form R sin(θ + α), where R > 0 and 0 < α < rac{π}{2}. Hence, for 0 ≤ θ < π, solve 2 sin 2θ = rac{3}{2}(cos 2θ + sin θ - 1), giving your answers in radians to 3 significant figures, where appropriate.

A-Level Edexcel Maths Pure Exponential & Logarithms: Using the identity cos(A + B) =

Using the identity cos(A + B) = cos A cos B - sin A sin B, prove that cos 2A = 1 - 2 sin² A - Edexcel - A-Level Maths Pure - Question 7 - 2005 - Paper 5

Question 7

Using the identity cos(A + B) = cos A cos B - sin A sin B, prove that cos 2A = 1 - 2 sin² A. Show that 2 sin 2θ - 3 cos 2θ - 3 sin θ + 3 = sin θ(4 cos θ + 6 sin θ -... show full transcript

Worked Solution & Example Answer:Using the identity cos(A + B) = cos A cos B - sin A sin B, prove that cos 2A = 1 - 2 sin² A - Edexcel - A-Level Maths Pure - Question 7 - 2005 - Paper 5

Step 1

Using the identity cos(A + B) = cos A cos B - sin A sin B, prove that cos 2A = 1 - 2 sin² A.

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Answer

To prove this, start with the identity for cosine:

egin{align*} ext{cos}(2A) &= ext{cos}(A + A) \ &= ext{cos} A ext{cos} A - ext{sin} A ext{sin} A \ &= ext{cos}^2 A - ext{sin}^2 A \ \ &= 1 - ext{sin}^2 A - ext{sin}^2 A \ &= 1 - 2 ext{sin}^2 A. ext{This completes the proof.} \end{align*}

Step 2

Show that 2 sin 2θ - 3 cos 2θ - 3 sin θ + 3 = sin θ(4 cos θ + 6 sin θ - 3).

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Answer

We start with the left-hand side:

egin{align*} 2 ext{sin}(2 heta) - 3 ext{cos}(2 heta) - 3 ext{sin} heta + 3 &= 2(2 ext{sin} heta ext{cos} heta) - 3(1 - 2 ext{sin}^2 heta) - 3 ext{sin} heta + 3 \ &= 4 ext{sin} heta ext{cos} heta - 3 + 6 ext{sin}^2 heta - 3 ext{sin} heta + 3 \ &= 4 ext{sin} heta ext{cos} heta + 6 ext{sin}^2 heta - 3 ext{sin} heta.\ \ ext{Factor out } ext{sin} heta: \ &= ext{sin} heta(4 ext{cos} heta + 6 ext{sin} heta - 3). ext{This establishes the equality.} \end{align*}

Step 3

Express 4 cos θ + 6 sin θ in the form R sin(θ + α), where R > 0 and 0 < α < π/2.

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Answer

To express this in the required form, we start by determining R and α:

Calculate R:

$R = ext{sqrt}(4^2 + 6^2) = ext{sqrt}(16 + 36) = ext{sqrt}(52) = 2 ext{sqrt}(13) \approx 7.21.$

Find α using:

α = ext{tan}^{-1}(1.5) \approx 0.588.$$ Thus, the expression becomes: $$ 4 ext{cos} θ + 6 ext{sin} θ = R ext{sin}(θ + α). $$

Step 4

Hence, for 0 ≤ θ < π, solve 2 sin 2θ = rac{3}{2}(cos 2θ + sin θ - 1).

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Answer

From the previous steps, we have:

Rewrite the equation:

$2 ext{sin}(2 heta) = rac{3}{2}( ext{cos}(2 heta) + ext{sin} heta - 1).$

Solve for θ:

Using numerical methods or graphing techniques, we find:

$heta ext{ values are: } 2.12, 0.588, ext{and others under given constraints.}$

The solutions in radians to 3 significant figures are:

$heta ext{ values: } 2.12, 0.588.$

Using the identity cos(A + B) = cos A cos B - sin A sin B, prove that cos 2A = 1 - 2 sin² A - Edexcel - A-Level Maths Pure - Question 7 - 2005 - Paper 5

Worked Solution & Example Answer:Using the identity cos(A + B) = cos A cos B - sin A sin B, prove that cos 2A = 1 - 2 sin² A - Edexcel - A-Level Maths Pure - Question 7 - 2005 - Paper 5

Using the identity cos(A + B) = cos A cos B - sin A sin B, prove that cos 2A = 1 - 2 sin² A.

Show that 2 sin 2θ - 3 cos 2θ - 3 sin θ + 3 = sin θ(4 cos θ + 6 sin θ - 3).

Express 4 cos θ + 6 sin θ in the form R sin(θ + α), where R > 0 and 0 < α < π/2.

Hence, for 0 ≤ θ < π, solve 2 sin 2θ = rac{3}{2}(cos 2θ + sin θ - 1).

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