Hard A-Level Compound and double angles Questions

Challenging, exam-style A-Level Compound and double angles questions with worked solutions. Stretch yourself on the hardest double angle, trig equations, addition formulae, compound angles problems.

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A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
A student proves cos⁡3x≡4cos⁡3x−3cos⁡x\cos 3x\equiv4\cos^3 x-3\cos x. Which statement best describes the result?
Show worked solution

Worked solution

  1. State the identity to be proved

    cos⁡3x≡4cos⁡3x−3cos⁡x\cos 3x\equiv 4\cos^3 x-3\cos x

    We work on one side until it matches the other.

  2. Write cos⁡3x=cos⁡(2x+x)\cos 3x=\cos(2x+x)

    =cos⁡2xcos⁡x−sin⁡2xsin⁡x=\cos 2x\cos x-\sin 2x\sin x

    Apply the cosine addition formula.

  3. Substitute cos⁡2x=2cos⁡2x−1\cos 2x=2\cos^2 x-1 and sin⁡2x=2sin⁡xcos⁡x\sin 2x=2\sin x\cos x

    =(2cos⁡2x−1)cos⁡x−2sin⁡2xcos⁡x=(2\cos^2 x-1)\cos x-2\sin^2 x\cos x

    Use both double-angle identities.

  4. Replace sin⁡2x\sin^2 x with 1−cos⁡2x1-\cos^2 x

    =(2cos⁡2x−1)cos⁡x−2(1−cos⁡2x)cos⁡x=(2\cos^2 x-1)\cos x-2(1-\cos^2 x)\cos x

    Eliminate sine to leave a cosine cubic.

  5. Expand and collect terms

    =4cos⁡3x−3cos⁡x=4\cos^3 x-3\cos x

    Multiply out and simplify.

  6. Confirm the two sides are now identical

    cos⁡3x≡4cos⁡3x−3cos⁡x\cos 3x\equiv 4\cos^3 x-3\cos x

    The left-hand side has been transformed into the right-hand side.

  7. Check the identity numerically at a first test angle

    x=30∘: LHS=RHS=0x=30^{\circ}:\ \text{LHS}=\text{RHS}=0

    A numerical check gives confidence the manipulation is correct.

  8. Check the identity numerically at a second test angle

    x=60∘: LHS=RHS=−1x=60^{\circ}:\ \text{LHS}=\text{RHS}=-1

    Agreement at another angle supports the algebraic proof.

  9. State that the identity holds for all permissible x

    true for all valid x\text{true for all valid }x

    Since the algebra used only standard identities, it holds generally.

  10. Summarise the key manipulation used

    cos⁡3x→4cos⁡3x−3cos⁡x\cos 3x\rightarrow 4\cos^3 x-3\cos x

    The double-angle identity was the essential step.

  11. Note that only standard identities were used

    double-angle and Pythagorean identities\text{double-angle and Pythagorean identities}

    No unproven results are assumed anywhere in the argument.

  12. Restate the expression we started from

    cos⁡3x\cos 3x

    This was the left-hand side before any manipulation.

  13. Restate the expression we finished with

    4cos⁡3x−3cos⁡x4\cos^3 x-3\cos x

    This is the target right-hand side.

  14. Explain why no exceptional cases are missed

    excluding points where a denominator is 0\text{excluding points where a denominator is }0

    The identity is valid wherever both sides are defined.

  15. Describe what the completed proof shows

    The triple-angle identity cos⁡3x≡4cos⁡3x−3cos⁡x\cos 3x\equiv4\cos^3 x-3\cos x is established.

    The proof establishes the stated equivalence for all valid angles.

Answer
The triple-angle identity cos⁡3x≡4cos⁡3x−3cos⁡x\cos 3x\equiv4\cos^3 x-3\cos x is established.
Question 2
8 markschallenging
A student proves sin⁡2x1+cos⁡2x≡tan⁡x\dfrac{\sin 2x}{1+\cos 2x}\equiv\tan x. Which statement best describes the result?
Show worked solution

Worked solution

  1. State the identity to be proved

    sin⁡2x1+cos⁡2x≡tan⁡x\dfrac{\sin 2x}{1+\cos 2x}\equiv \tan x

    We work on one side until it matches the other.

  2. Replace sin⁡2x\sin 2x with 2sin⁡xcos⁡x2\sin x\cos x

    =2sin⁡xcos⁡x1+cos⁡2x=\dfrac{2\sin x\cos x}{1+\cos 2x}

    Use the sine double-angle identity in the numerator.

  3. Replace 1+cos⁡2x1+\cos 2x with 2cos⁡2x2\cos^2 x

    =2sin⁡xcos⁡x2cos⁡2x=\dfrac{2\sin x\cos x}{2\cos^2 x}

    Use the cosine-only double-angle form.

  4. Cancel 2cos⁡x2\cos x

    =sin⁡xcos⁡x=\dfrac{\sin x}{\cos x}

    Cancel common factors.

  5. Recognise the tangent ratio

    =tan⁡x=\tan x

    sin/cos is tan.

  6. Confirm the two sides are now identical

    sin⁡2x1+cos⁡2x≡tan⁡x\dfrac{\sin 2x}{1+\cos 2x}\equiv \tan x

    The left-hand side has been transformed into the right-hand side.

  7. Check the identity numerically at a first test angle

    x=30∘: LHS=RHS=33x=30^{\circ}:\ \text{LHS}=\text{RHS}=\frac{\sqrt{3}}{3}

    A numerical check gives confidence the manipulation is correct.

  8. Check the identity numerically at a second test angle

    x=60∘: LHS=RHS=3x=60^{\circ}:\ \text{LHS}=\text{RHS}=\sqrt{3}

    Agreement at another angle supports the algebraic proof.

  9. State that the identity holds for all permissible x

    true for all valid x\text{true for all valid }x

    Since the algebra used only standard identities, it holds generally.

  10. Summarise the key manipulation used

    sin⁡2x1+cos⁡2x→tan⁡x\dfrac{\sin 2x}{1+\cos 2x}\rightarrow \tan x

    The double-angle identity was the essential step.

  11. Note that only standard identities were used

    double-angle and Pythagorean identities\text{double-angle and Pythagorean identities}

    No unproven results are assumed anywhere in the argument.

  12. Restate the expression we started from

    sin⁡2x1+cos⁡2x\dfrac{\sin 2x}{1+\cos 2x}

    This was the left-hand side before any manipulation.

  13. Restate the expression we finished with

    tan⁡x\tan x

    This is the target right-hand side.

  14. Explain why no exceptional cases are missed

    excluding points where a denominator is 0\text{excluding points where a denominator is }0

    The identity is valid wherever both sides are defined.

  15. Describe what the completed proof shows

    Both sides are equal for all xx where cos⁡x≠0\cos x\neq0.

    The proof establishes the stated equivalence for all valid angles.

Answer
Both sides are equal for all xx where cos⁡x≠0\cos x\neq0.
Question 3
8 markschallenging
A student simplifies 2sin⁡xcos⁡x(cos⁡2x−sin⁡2x)2\sin x\cos x(\cos^2 x-\sin^2 x). Which statement best describes the result?
Show worked solution

Worked solution

  1. State the identity to be proved

    2sin⁡xcos⁡x(cos⁡2x−sin⁡2x)≡12sin⁡4x2\sin x\cos x(\cos^2 x-\sin^2 x)\equiv \tfrac{1}{2}\sin 4x

    We work on one side until it matches the other.

  2. Recognise 2sin⁡xcos⁡x=sin⁡2x2\sin x\cos x=\sin 2x

    =sin⁡2x(cos⁡2x−sin⁡2x)=\sin 2x(\cos^2 x-\sin^2 x)

    Apply the sine double-angle identity.

  3. Recognise cos⁡2x−sin⁡2x=cos⁡2x\cos^2 x-\sin^2 x=\cos 2x

    =sin⁡2xcos⁡2x=\sin 2x\cos 2x

    Apply the cosine double-angle identity.

  4. Use 2sin⁡2xcos⁡2x=sin⁡4x2\sin 2x\cos 2x=\sin 4x

    =12sin⁡4x=\tfrac{1}{2}\sin 4x

    Apply the double-angle identity again with angle 2x.

  5. Confirm the two sides are now identical

    2sin⁡xcos⁡x(cos⁡2x−sin⁡2x)≡12sin⁡4x2\sin x\cos x(\cos^2 x-\sin^2 x)\equiv \tfrac{1}{2}\sin 4x

    The left-hand side has been transformed into the right-hand side.

  6. Check the identity numerically at a first test angle

    x=30∘: LHS=RHS=34x=30^{\circ}:\ \text{LHS}=\text{RHS}=\frac{\sqrt{3}}{4}

    A numerical check gives confidence the manipulation is correct.

  7. Check the identity numerically at a second test angle

    x=60∘: LHS=RHS=−34x=60^{\circ}:\ \text{LHS}=\text{RHS}=- \frac{\sqrt{3}}{4}

    Agreement at another angle supports the algebraic proof.

  8. State that the identity holds for all permissible x

    true for all valid x\text{true for all valid }x

    Since the algebra used only standard identities, it holds generally.

  9. Summarise the key manipulation used

    2sin⁡xcos⁡x(cos⁡2x−sin⁡2x)→12sin⁡4x2\sin x\cos x(\cos^2 x-\sin^2 x)\rightarrow \tfrac{1}{2}\sin 4x

    The double-angle identity was the essential step.

  10. Note that only standard identities were used

    double-angle and Pythagorean identities\text{double-angle and Pythagorean identities}

    No unproven results are assumed anywhere in the argument.

  11. Restate the expression we started from

    2sin⁡xcos⁡x(cos⁡2x−sin⁡2x)2\sin x\cos x(\cos^2 x-\sin^2 x)

    This was the left-hand side before any manipulation.

  12. Restate the expression we finished with

    12sin⁡4x\tfrac{1}{2}\sin 4x

    This is the target right-hand side.

  13. Explain why no exceptional cases are missed

    excluding points where a denominator is 0\text{excluding points where a denominator is }0

    The identity is valid wherever both sides are defined.

  14. Conclude the proof is complete

    ■\blacksquare

    The left-hand side equals the right-hand side as required.

  15. Describe what the completed proof shows

    Repeated use of the double-angle formulae gives 12sin⁡4x\tfrac{1}{2}\sin 4x.

    The proof establishes the stated equivalence for all valid angles.

Answer
Repeated use of the double-angle formulae gives 12sin⁡4x\tfrac{1}{2}\sin 4x.
Question 4
8 markschallenging
A student proves cos⁡4x−sin⁡4x≡cos⁡2x\cos^4 x-\sin^4 x\equiv\cos 2x. Which statement best describes the result?
Show worked solution

Worked solution

  1. State the identity to be proved

    cos⁡4x−sin⁡4x≡cos⁡2x\cos^4 x-\sin^4 x\equiv \cos 2x

    We work on one side until it matches the other.

  2. Factorise as a difference of two squares

    =(cos⁡2x−sin⁡2x)(cos⁡2x+sin⁡2x)=(\cos^2 x-\sin^2 x)(\cos^2 x+\sin^2 x)

    a^2-b^2=(a-b)(a+b).

  3. Use cos⁡2x+sin⁡2x≡1\cos^2 x+\sin^2 x\equiv1

    =cos⁡2x−sin⁡2x=\cos^2 x-\sin^2 x

    The second bracket equals 1.

  4. Recognise the double-angle identity

    =cos⁡2x=\cos 2x

    cos^2 - sin^2 is cos 2x.

  5. Confirm the two sides are now identical

    cos⁡4x−sin⁡4x≡cos⁡2x\cos^4 x-\sin^4 x\equiv \cos 2x

    The left-hand side has been transformed into the right-hand side.

  6. Check the identity numerically at a first test angle

    x=30∘: LHS=RHS=12x=30^{\circ}:\ \text{LHS}=\text{RHS}=\frac{1}{2}

    A numerical check gives confidence the manipulation is correct.

  7. Check the identity numerically at a second test angle

    x=60∘: LHS=RHS=−12x=60^{\circ}:\ \text{LHS}=\text{RHS}=- \frac{1}{2}

    Agreement at another angle supports the algebraic proof.

  8. State that the identity holds for all permissible x

    true for all valid x\text{true for all valid }x

    Since the algebra used only standard identities, it holds generally.

  9. Summarise the key manipulation used

    cos⁡4x−sin⁡4x→cos⁡2x\cos^4 x-\sin^4 x\rightarrow \cos 2x

    The double-angle identity was the essential step.

  10. Note that only standard identities were used

    double-angle and Pythagorean identities\text{double-angle and Pythagorean identities}

    No unproven results are assumed anywhere in the argument.

  11. Restate the expression we started from

    cos⁡4x−sin⁡4x\cos^4 x-\sin^4 x

    This was the left-hand side before any manipulation.

  12. Restate the expression we finished with

    cos⁡2x\cos 2x

    This is the target right-hand side.

  13. Explain why no exceptional cases are missed

    excluding points where a denominator is 0\text{excluding points where a denominator is }0

    The identity is valid wherever both sides are defined.

  14. Conclude the proof is complete

    ■\blacksquare

    The left-hand side equals the right-hand side as required.

  15. Describe what the completed proof shows

    The expression simplifies exactly to cos⁡2x\cos 2x for all xx.

    The proof establishes the stated equivalence for all valid angles.

Answer
The expression simplifies exactly to cos⁡2x\cos 2x for all xx.
Question 5
8 markschallenging
A student proves sin⁡2x1−cos⁡2x≡1tan⁡x\dfrac{\sin 2x}{1-\cos 2x}\equiv\dfrac{1}{\tan x}. Which statement best describes the result?
Show worked solution

Worked solution

  1. State the identity to be proved

    sin⁡2x1−cos⁡2x≡1tan⁡x\dfrac{\sin 2x}{1-\cos 2x}\equiv \dfrac{1}{\tan x}

    We work on one side until it matches the other.

  2. Replace sin⁡2x\sin 2x with 2sin⁡xcos⁡x2\sin x\cos x

    =2sin⁡xcos⁡x1−cos⁡2x=\dfrac{2\sin x\cos x}{1-\cos 2x}

    Use the sine double-angle identity in the numerator.

  3. Replace 1−cos⁡2x1-\cos 2x with 2sin⁡2x2\sin^2 x

    =2sin⁡xcos⁡x2sin⁡2x=\dfrac{2\sin x\cos x}{2\sin^2 x}

    Use the sine-only double-angle form.

  4. Cancel 2sin⁡x2\sin x

    =cos⁡xsin⁡x=\dfrac{\cos x}{\sin x}

    Cancel common factors.

  5. Recognise the reciprocal tangent

    =1tan⁡x=\dfrac{1}{\tan x}

    cos/sin is the reciprocal of tan.

  6. Confirm the two sides are now identical

    sin⁡2x1−cos⁡2x≡1tan⁡x\dfrac{\sin 2x}{1-\cos 2x}\equiv \dfrac{1}{\tan x}

    The left-hand side has been transformed into the right-hand side.

  7. Check the identity numerically at a first test angle

    x=30∘: LHS=RHS=3x=30^{\circ}:\ \text{LHS}=\text{RHS}=\sqrt{3}

    A numerical check gives confidence the manipulation is correct.

  8. Check the identity numerically at a second test angle

    x=60∘: LHS=RHS=33x=60^{\circ}:\ \text{LHS}=\text{RHS}=\frac{\sqrt{3}}{3}

    Agreement at another angle supports the algebraic proof.

  9. State that the identity holds for all permissible x

    true for all valid x\text{true for all valid }x

    Since the algebra used only standard identities, it holds generally.

  10. Summarise the key manipulation used

    sin⁡2x1−cos⁡2x→1tan⁡x\dfrac{\sin 2x}{1-\cos 2x}\rightarrow \dfrac{1}{\tan x}

    The double-angle identity was the essential step.

  11. Note that only standard identities were used

    double-angle and Pythagorean identities\text{double-angle and Pythagorean identities}

    No unproven results are assumed anywhere in the argument.

  12. Restate the expression we started from

    sin⁡2x1−cos⁡2x\dfrac{\sin 2x}{1-\cos 2x}

    This was the left-hand side before any manipulation.

  13. Restate the expression we finished with

    1tan⁡x\dfrac{1}{\tan x}

    This is the target right-hand side.

  14. Explain why no exceptional cases are missed

    excluding points where a denominator is 0\text{excluding points where a denominator is }0

    The identity is valid wherever both sides are defined.

  15. Describe what the completed proof shows

    Both sides are equal for all xx where sin⁡x≠0\sin x\neq0.

    The proof establishes the stated equivalence for all valid angles.

Answer
Both sides are equal for all xx where sin⁡x≠0\sin x\neq0.

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