A-Level Compound and double angles Practice Questions

Free A-Level Compound and double angles practice questions with full step-by-step worked solutions. Covers addition formulae, exact values, double angle, sin 2x. Practise exam-style problems and check your method.

addition formulaeexact valuesdouble anglesin 2xcos 2xcompound angles
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
Using an addition formula, find the exact value of sin⁡75∘\sin 75^{\circ}.
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Worked solution

  1. Write the angle as a sum or difference of two special angles

    75∘=45∘+30∘75^{\circ}=45^{\circ}+30^{\circ}

    Choose familiar angles (30, 45, 60 degrees) whose exact ratios are known.

  2. State the relevant addition formula

    sin⁡(A+B)≡sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B)\equiv \sin A\cos B+\cos A\sin B

    The compound-angle formula rewrites the ratio of a sum in terms of the parts.

  3. State the exact value

    sin⁡75∘=24+64\sin 75^{\circ}=\frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4}

    This is the required exact value.

Answer
24+64\frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4}
Question 2
2 markseasy
Which expression is equal to cos⁡2x\cos 2x?
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Worked solution

  1. Recall the standard formula

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

    This is the identity being tested.

  2. Compare the term structure with each option

    match signs and factors\text{match signs and factors}

    The correct option must have the exact same terms and signs.

  3. Select the correct statement

    cos⁡2x−sin⁡2x\cos^2 x-\sin^2 x

    Only this option matches the standard identity.

Answer
cos⁡2x−sin⁡2x\cos^2 x-\sin^2 x
Question 3
3 marksintermediate
Which is the correct formula for tan⁡(A−B)\tan(A-B)?
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Worked solution

  1. Recall the tangent addition formula

    tan⁡(A+B)≡tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan(A+B)\equiv\dfrac{\tan A+\tan B}{1-\tan A\tan B}

    Start from the sum formula.

  2. Replace B by -B

    tan⁡(A−B)=tan⁡(A+(−B))\tan(A-B)=\tan(A+(-B))

    A difference is a sum with a negative angle.

  3. Use tan⁡(−B)=−tan⁡B\tan(-B)=-\tan B

    =tan⁡A−tan⁡B1−tan⁡A(−tan⁡B)=\dfrac{\tan A-\tan B}{1-\tan A(-\tan B)}

    Tangent is an odd function.

  4. Simplify the denominator

    =tan⁡A−tan⁡B1+tan⁡Atan⁡B=\dfrac{\tan A-\tan B}{1+\tan A\tan B}

    The double negative becomes a plus.

  5. State the formula

    tan⁡(A−B)≡tan⁡A−tan⁡B1+tan⁡Atan⁡B\tan(A-B)\equiv\dfrac{\tan A-\tan B}{1+\tan A\tan B}

    This is the standard difference formula.

  6. Select the equivalent expression

    tan⁡A−tan⁡B1+tan⁡Atan⁡B\dfrac{\tan A-\tan B}{1+\tan A\tan B}

    This form follows directly from the derivation above.

Answer
tan⁡A−tan⁡B1+tan⁡Atan⁡B\dfrac{\tan A-\tan B}{1+\tan A\tan B}
Question 4
5 markshard
In proving sin⁡2x1+cos⁡2x≡tan⁡x\dfrac{\sin 2x}{1+\cos 2x}\equiv\tan x, which identity is used in the numerator?
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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 double-angle identity in the denominator.

  4. Cancel 2cos⁡x2\cos x

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

    Cancel the 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. Identify the key identity used in the proof

    sin⁡2x≡2sin⁡xcos⁡x\sin 2x\equiv2\sin x\cos x

    This double-angle identity is what makes the proof work.

Answer
sin⁡2x≡2sin⁡xcos⁡x\sin 2x\equiv2\sin x\cos x
Question 5
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?
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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.

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