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Next exam on November 14 – 15, 2026:
All concepts

Double-Angle Formulas

二倍角公式

Double-angle formulas for sine, cosine and tangent; the three equivalent forms of cos 2α do most of the simplification work.

5 real-exam questions · 5 sittings22 in mock papers6 practice

Formulas

  • Double-Angle Sine

    Identity
    sin⁡2α=2sin⁡αcos⁡α\sin 2\alpha=2\sin\alpha\cos\alpha
  • Double-Angle Cosine

    Identity
    cos⁡2α=cos⁡2α−sin⁡2α=2cos⁡2α−1=1−2sin⁡2α\cos 2\alpha=\cos^{2}\alpha-\sin^{2}\alpha=2\cos^{2}\alpha-1=1-2\sin^{2}\alpha
  • Double-Angle Tangent

    Identity
    tan⁡2α=2tan⁡α1−tan⁡2α\tan 2\alpha=\frac{2\tan\alpha}{1-\tan^{2}\alpha}

    Applies when:tan α defined and tan²α ≠ 1

How the real exam tests it

Questions from real CSCA papers that test this concept. Work each one out before you open the answer.

  1. CSCA Math December 2025 Past Paper
    If sin⁡α=14\sin\alpha=\frac{1}{4}, then cos⁡2α=\cos2\alpha= ( ).
    • A.−78-\frac{7}{8}
    • B.18\frac{1}{8}
    • C.1516\frac{15}{16}
    • D.78\frac{7}{8}
    Show answer & solution

    Answer: D

    Answer: D cos⁡2α=1−2sin⁡2α=1−2⋅116=78\cos2\alpha=1-2\sin^2\alpha=1-2\cdot\frac{1}{16}=\frac{7}{8}. Choose D.
    See it in the full CSCA Math December 2025 Past Paper
  2. CSCA Math January 2026 Past Paper
    If sin⁡α=35\sin \alpha = \frac { 3 } { 5 }, and α\alpha belongs to the first quadrant, then sin⁡2α=()\sin 2 \alpha = ( ).
    • A.1225\frac { 12 } { 25 }
    • B.2425\frac { 24 } { 25 }
    • C.1825\frac { 18 } { 25 }
    • D.725\frac { 7 } { 25 }
    Show answer & solution

    Answer: B

    Given sin⁡α=3/5\sin \alpha = 3 / 5 and α\alpha is in Quadrant I. Find sin⁡2α\sin 2 \alpha. In Quadrant I, cos⁡α>0\cos \alpha > 0. cos⁡α=1−sin⁡2α=1−9/25=16/25=4/5\cos \alpha = \sqrt { 1 - \sin ^ { 2 } \alpha } = \sqrt { 1 - 9 / 25 } = \sqrt { 16 / 25 } = 4 / 5. sin⁡2α=2sin⁡αcos⁡α=2⋅(3/5)⋅(4/5)=24/25\sin 2 \alpha = 2 \sin \alpha \cos \alpha = 2 \cdot ( 3 / 5 ) \cdot ( 4 / 5 ) = 24 / 25.
    Take the full CSCA Math January 2026 Past Paper — timed & graded
  3. CSCA Math March 2026 Past Paper
    If sin⁡α=3/3\sin \alpha = \sqrt { 3 } / 3, then cos⁡2α=\cos 2 \alpha =
    • A.2/32 / 3
    • B.1/31 / 3
    • C.−1/3- 1 / 3
    • D.−2/3- 2 / 3
    Show answer & solution

    Answer: B

    Double angle formula: cos⁡2α=1−2sin⁡2α=1−2×(1/3)=1/3\cos 2 \alpha = 1 - 2 \sin ^ { 2 } \alpha = 1 - 2 \times ( 1 / 3 ) = 1 / 3.
    Take the full CSCA Math March 2026 Past Paper — timed & graded

Real-exam evidence

Recorded past-paper appearances are listed below; online verification is labelled separately.

5 questions across 5 sittings1 of 5 sittings available to verify free online
  • CSCA Math June 2026 Past Paper1 questionreading page not yet public
  • CSCA Math April 2026 Past Paper1 questionreading page not yet public
  • CSCA Math March 2026 Past Paper1 questionreading page not yet public
  • CSCA Math January 2026 Past Paper1 questionreading page not yet public
  • CSCA Math December 2025 Past Paper1 questionView question 27 and answer free (full paper)

22 questions across 12 mock papers — our own papers, so a coverage signal rather than real-exam evidence

6 related practice questions availablePractise free: Elementary Function

Sitting evidence comes from the CSCA past papers on record; questions whose original paper is missing are excluded, and sprint papers are deliberately not counted.

Glossary

Related concepts