例題 1 Example 1：求其他三角比 Find the other ratios
sin θ = 3/5，θ 為銳角，求 cos θ 及 tan θ。sin θ = 3/5 and θ is acute. Find cos θ and tan θ.
cos²θ = 1 − 9/25 = 16/25- θ 是銳角，所以
cos θ > 0θ is acute, socos θ > 0
答案： Answer: cos θ = 4/5, tan θ = 3/4
【三角函數及恆等式：sin²θ + cos²θ = 1 及 tan θ = sin θ / cos θ】Trigonometric Ratios and Identities: sin²θ + cos²θ = 1 and tan θ = sin θ / cos θ
SketchMath 數學團隊 SketchMath Maths Team ｜ 更新 Updated 2026-10-20
對任何角 θ：sin²θ + cos²θ = 1，tan θ = sin θ / cos θ。在單位圓上，角 θ 對應的點是 (cos θ, sin θ)。解 0° ≤ θ < 360° 的方程時，先求銳角參考角，再按象限的正負號（All、Sin、Tan、Cos）決定答案。For any angle θ: sin²θ + cos²θ = 1 and tan θ = sin θ / cos θ. On the unit circle the point at angle θ is (cos θ, sin θ). To solve 0° ≤ θ < 360°, find the acute reference angle, then use the quadrant signs (All, Sin, Tan, Cos).
三角函數恆等式 Trigonometric Identities
畢氏恆等式Pythagorean identity
sin²θ + cos²θ = 1正切Tangent
tan θ = sin θ / cos θ變形Rearranged forms
sin²θ = 1 − cos²θ, cos²θ = 1 − sin²θ象限的正負號Quadrant signs第一象限：全部為正；第二象限：sin；第三象限：tan；第四象限：cos1st: all positive; 2nd: sin; 3rd: tan; 4th: cos (ASTC)
特殊角Special angles
sin 30° = ½, cos 60° = ½, sin 45° = cos 45° = √2/2, sin 60° = cos 30° = √3/2sin、cos、tan 的正負號：ASTC The Signs of sin, cos and tan: ASTC
由第一象限逆時針方向，三角比為正的依次是 All（全部）、Sin、Tan、Cos。解 0° ≤ θ < 360° 時，先求銳角參考角，再按這個正負號決定答案。
Going anticlockwise from the first quadrant, the ratios that are positive are All, Sin, Tan, Cos. To solve 0° ≤ θ < 360°, find the acute reference angle and then use these signs.
三角函數及恆等式例題詳解 Worked Examples
例題 2 Example 2：用恆等式化簡 Simplify using an identity
化簡 sin²θ / (1 − cos θ)。Simplify sin²θ / (1 − cos θ).
sin²θ = 1 − cos²θ = (1 − cos θ)(1 + cos θ)- 約去
1 − cos θ（cos θ ≠ 1）Cancel1 − cos θ(cos θ ≠ 1)
答案： Answer: 1 + cos θ
例題 3 Example 3：解三角方程 Solve a trigonometric equation
解 2sin²θ = 1 + cos θ，0° ≤ θ < 360°。Solve 2sin²θ = 1 + cos θ for 0° ≤ θ < 360°.
2(1 − cos²θ) = 1 + cos θ, so2cos²θ + cos θ − 1 = 0(2cos θ − 1)(cos θ + 1) = 0, socos θ = 1/2orcos θ = −1cos θ = 1/2：θ = 60°, 300°；cos θ = −1：θ = 180°cos θ = 1/2:θ = 60°, 300°;cos θ = −1:θ = 180°
答案： Answer: θ = 60°, 180°, 300°
💡 應試陷阱與秘訣 Exam Traps & Secrets
⚠️ 不要漏掉負根 Do not lose the negative root由
cos²θ = 16/25 得 cos θ = ± 4/5。要看 θ 所在象限來決定取哪個符號。From cos²θ = 16/25, cos θ = ± 4/5. The quadrant of θ decides which sign you keep.⚠️ 不要除以三角函數 Never divide by a trigonometric factor兩邊同除以
cos θ 可能漏掉 cos θ = 0 的根。應把所有項移到一邊再分解因式。Dividing both sides by cos θ can lose the roots where cos θ = 0. Move everything to one side and factorize instead.✏️ 自我測試：互動練習 Self-Test: Interactive Practice
溫習方法：點擊以下任何題目，即時在坐標網格載入並核對計算步驟。How to study: click any question to load it on the coordinate grid and check your steps.
(證明 Prove)證明 (1 − sin θ)(1 + sin θ) = cos²θ。Prove that (1 − sin θ)(1 + sin θ) = cos²θ.
(解方程 Solve)解 2cos²θ − sin θ − 1 = 0，0° ≤ θ < 360°。Solve 2cos²θ − sin θ − 1 = 0 for 0° ≤ θ < 360°.
(象限 Quadrants)已知 tan θ = 5/12，θ 在第三象限，求 sin θ 及 cos θ。Given tan θ = 5/12 and θ in the third quadrant, find sin θ and cos θ.
三角函數及恆等式練習題（DSE 模擬題） Exercises: DSE-Style Practice
SketchMath 題庫內的 DSE 風格練習，選擇題及長題目均附解答。 DSE-style practice from the SketchMath question bank, with solutions for every MC and long question.
三角比的關係 · 中三
31 條選擇題 31 MC · 25 條長題目 25 long questions
三角函數、圖像及方程 · 中四
113 條選擇題 113 MC · 16 條長題目 16 long questions
常見問題 Frequently Asked Questions
在單位圓上，點 (cos θ, sin θ) 與原點相距 1，由畢氏定理得 cos²θ + sin²θ = 1。
On the unit circle the point (cos θ, sin θ) is 1 unit from the origin, so cos²θ + sin²θ = 1 by Pythagoras.
它說明每個象限哪個三角比為正：由第一象限逆時針依次是 All、Sin、Tan、Cos。
It lists which ratio is positive in each quadrant: All, Sin, Tan, Cos, going anticlockwise from the first quadrant.
由較複雜的一邊開始，用恆等式化簡，直至與另一邊相同。不要移項。
Start from the more complicated side and simplify it using the identities until it matches the other side. Do not move terms across.