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三角恒等式sānjiǎo héngděngshì

trigonometric identities
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更新于 2025-11-02
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Core Concept

Trigonometric Identities are equations involving trigonometric functions that are true for all values in their domain. Mastering these identities is key to learning trigonometry.

Fundamental Identities

Pythagorean Identities

sin2α+cos2α=1\sin^2\alpha + \cos^2\alpha = 1 1+tan2α=sec2α1 + \tan^2\alpha = \sec^2\alpha

Quotient Identities

tanα=sinαcosα\tan\alpha = \frac{\sin\alpha}{\cos\alpha}

Reduction Formulas

Negative Angles

sin(α)=sinα\sin(-\alpha) = -\sin\alpha cos(α)=cosα\cos(-\alpha) = \cos\alpha

Complementary Angles

sin(π2α)=cosα\sin\left(\frac{\pi}{2} - \alpha\right) = \cos\alpha cos(π2α)=sinα\cos\left(\frac{\pi}{2} - \alpha\right) = \sin\alpha

Supplementary Angles

sin(πα)=sinα\sin(\pi - \alpha) = \sin\alpha cos(πα)=cosα\cos(\pi - \alpha) = -\cos\alpha

Sum and Difference Formulas

Sine

sin(α±β)=sinαcosβ±cosαsinβ\sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \cos\alpha\sin\beta

Cosine

cos(α±β)=cosαcosβsinαsinβ\cos(\alpha \pm \beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta

Tangent

tan(α±β)=tanα±tanβ1tanαtanβ\tan(\alpha \pm \beta) = \frac{\tan\alpha \pm \tan\beta}{1 \mp \tan\alpha\tan\beta}

Double Angle Formulas

Sine

sin2α=2sinαcosα\sin 2\alpha = 2\sin\alpha\cos\alpha

Cosine

cos2α=cos2αsin2α=2cos2α1=12sin2α\cos 2\alpha = \cos^2\alpha - \sin^2\alpha = 2\cos^2\alpha - 1 = 1 - 2\sin^2\alpha

Power-Reducing Formulas: cos2α=1+cos2α2\cos^2\alpha = \frac{1 + \cos 2\alpha}{2} sin2α=1cos2α2\sin^2\alpha = \frac{1 - \cos 2\alpha}{2}

Tangent

tan2α=2tanα1tan2α\tan 2\alpha = \frac{2\tan\alpha}{1 - \tan^2\alpha}

Auxiliary Angle Formula

asinx+bcosx=a2+b2sin(x+φ)a\sin x + b\cos x = \sqrt{a^2 + b^2}\sin(x + \varphi)

where tanφ=ba\tan\varphi = \frac{b}{a}

CSCA Practice Problems

[Example 1] Basic (Difficulty ★★☆☆☆)

Simplify: sin2α+cos2α+tan2α\sin^2\alpha + \cos^2\alpha + \tan^2\alpha

Solution: sin2α+cos2α+tan2α=1+tan2α=sec2α\sin^2\alpha + \cos^2\alpha + \tan^2\alpha = 1 + \tan^2\alpha = \sec^2\alpha

Answer: sec2α\sec^2\alpha


[Example 2] Intermediate (Difficulty ★★★☆☆)

Given sinα=35\sin\alpha = \frac{3}{5}, α(0,π2)\alpha \in (0, \frac{\pi}{2}), find sin2α\sin 2\alpha.

Solution:

cosα=1sin2α=45\cos\alpha = \sqrt{1 - \sin^2\alpha} = \frac{4}{5}

sin2α=2sinαcosα=2×35×45=2425\sin 2\alpha = 2\sin\alpha\cos\alpha = 2 \times \frac{3}{5} \times \frac{4}{5} = \frac{24}{25}

Answer: 2425\frac{24}{25}

Common Misconceptions

❌ Misconception 1: Wrong reduction formula

Wrong: sin(πα)=sinα\sin(\pi - \alpha) = -\sin\alpha

Correct: sin(πα)=sinα\sin(\pi - \alpha) = \sin\alpha

❌ Misconception 2: Confusing sum formulas

Wrong: sin(α+β)=sinα+sinβ\sin(\alpha + \beta) = \sin\alpha + \sin\beta

Correct: sin(α+β)=sinαcosβ+cosαsinβ\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta

Study Tips

  1. Categorize: Basic, reduction, sum/difference, double angle
  2. Understand derivation: Don't just memorize
  3. Practice: Reinforce through exercises
  4. Connect formulas: Many can be derived from each other

💡 Exam Tip: Trigonometric identities are core to trigonometry, mandatory in CSCA! Mastering formulas is key to problem-solving.