Trigonometric Identities and Equations三角恒等式与方程

Three identities, one geometric source.三条恒等式，同一几何来源。

Every point $(\cos\theta, \sin\theta)$ on the unit circle satisfies $x^{2} + y^{2} = 1$. That single fact powers all three identities below; divide it through by $\cos^{2}\theta$ or $\sin^{2}\theta$ to get the other two.单位圆上每个点 $(\cos\theta, \sin\theta)$ 都满足 $x^{2} + y^{2} = 1$。这一事实即为下面三条恒等式的来源；两边除以 $\cos^{2}\theta$ 或 $\sin^{2}\theta$ 即得另外两条。

$$ \sin^{2}\theta + \cos^{2}\theta = 1 $$ $$ 1 + \tan^{2}\theta = \sec^{2}\theta $$ $$ 1 + \cot^{2}\theta = \csc^{2}\theta $$

• Use when:使用时机： you know one of $\{\sin, \cos, \tan, \sec, \csc, \cot\}$ and the quadrant, and need another. CCSSM HSF-TF.C.8 names exactly this task.已知 $\{\sin, \cos, \tan, \sec, \csc, \cot\}$ 之一与象限，求其它。CCSSM HSF-TF.C.8 恰指此任务。
• Sign comes from the quadrant.符号取决于象限。 The identity gives you $\cos^{2}\theta$; take a square root and choose $+$ or $-$ from the quadrant (Q I and IV: $\cos > 0$; Q II and III: $\cos < 0$).恒等式给出 $\cos^{2}\theta$；开方后根据象限选符号（Q I、IV：$\cos > 0$；Q II、III：$\cos < 0$）。
• Reformulations.变形。 $\sin^{2}\theta = 1 - \cos^{2}\theta$, $\cos^{2}\theta = 1 - \sin^{2}\theta$, $\sec^{2}\theta - \tan^{2}\theta = 1$, $\csc^{2}\theta - \cot^{2}\theta = 1$. All four come up inside identity proofs.$\sin^{2}\theta = 1 - \cos^{2}\theta$、$\cos^{2}\theta = 1 - \sin^{2}\theta$、$\sec^{2}\theta - \tan^{2}\theta = 1$、$\csc^{2}\theta - \cot^{2}\theta = 1$。四种形式都会在恒等式证明中出现。

BC PC 12 lists "Pythagorean" first in its identity catalogue and AB Math 30-1 outcome 6 names it as the third bullet under "prove trigonometric identities using", both treat it as the foundation that the other identities are verified against.BC PC 12 把"勾股"列为恒等式清单首项；AB Math 30-1 目标 6 在"使用以下恒等式证明三角恒等式"中将其列为第三项 , 两套大纲都把它视为其他恒等式赖以检验的基础。

Six functions, two patterns.六个函数，两种模式。

Quotient identities , define $\tan$ and $\cot$ from $\sin$ and $\cos$:商恒等式 , 用 $\sin$ 与 $\cos$ 定义 $\tan$ 与 $\cot$：

$$ \tan\theta = \frac{\sin\theta}{\cos\theta}, \qquad \cot\theta = \frac{\cos\theta}{\sin\theta}. $$

Reciprocal identities , define $\sec, \csc, \cot$ as $1/\cos, 1/\sin, 1/\tan$:倒数恒等式 , 把 $\sec, \csc, \cot$ 定义为 $1/\cos, 1/\sin, 1/\tan$：

$$ \sec\theta = \frac{1}{\cos\theta}, \qquad \csc\theta = \frac{1}{\sin\theta}, \qquad \cot\theta = \frac{1}{\tan\theta}. $$

• Mnemonic.口诀。 $\sec$ pairs with $\cos$ (the one that starts with "c"); $\csc$ pairs with $\sin$ (the cross-pairing trips students up).$\sec$ 与 $\cos$ 配对（首字母"c"）；$\csc$ 与 $\sin$ 配对（这种交叉配对容易记错）。
• Non-permissible values (AB Math 30-1 indicator 6.5).非许可值（AB Math 30-1 指标 6.5）。 $\sec\theta$ undefined when $\cos\theta = 0$ (i.e. $\theta = \pi/2 + k\pi$). $\csc\theta$ undefined when $\sin\theta = 0$ ($\theta = k\pi$). $\tan\theta, \sec\theta$ share the same restriction; $\cot\theta, \csc\theta$ share theirs.当 $\cos\theta = 0$（即 $\theta = \pi/2 + k\pi$）时 $\sec\theta$ 无定义；当 $\sin\theta = 0$（$\theta = k\pi$）时 $\csc\theta$ 无定义。$\tan\theta$、$\sec\theta$ 限制相同；$\cot\theta$、$\csc\theta$ 限制相同。
• Strategy.策略。 When verifying an identity or solving an equation, convert everything to $\sin$ and $\cos$ first. This single move handles 80% of identity proofs in this unit.证明恒等式或解方程时，先把所有三角函数转写为 $\sin$ 与 $\cos$。本单元 80% 的恒等式证明靠这一步解决。

PC 12 Content elaboration names "quotient" and "reciprocal" alongside Pythagorean in the identity catalogue. AB Math 30-1 outcome 6 names them as the first two bullets under "prove trig identities using."PC 12 内容详释在恒等式清单中把"商"与"倒数"和勾股并列。AB Math 30-1 目标 6 在"使用以下恒等式证明三角恒等式"中将其列为前两项。

Six formulas, three patterns.六个公式，三种模式。 $$ \sin(A \pm B) = \sin A \cos B \pm \cos A \sin B $$ $$ \cos(A \pm B) = \cos A \cos B \mp \sin A \sin B $$ $$ \tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B} $$

• Sign rule for cosine.余弦符号规则。 The cosine formula flips the sign between the two terms: $\cos(A + B) = \cos A \cos B - \sin A \sin B$, $\cos(A - B) = \cos A \cos B + \sin A \sin B$. Memorise the master formula $\cos(A - B) = \cos A \cos B + \sin A \sin B$ and derive the rest.余弦公式中两项之间的符号反转：$\cos(A + B) = \cos A \cos B - \sin A \sin B$，$\cos(A - B) = \cos A \cos B + \sin A \sin B$。背熟主公式 $\cos(A - B) = \cos A \cos B + \sin A \sin B$，其余由此派生。
• Sign rule for sine.正弦符号规则。 The sign on $B$ in the argument matches the sign on the second term: $\sin(A + B)$ uses $+$, $\sin(A - B)$ uses $-$.参数中 $B$ 前的符号与第二项符号一致：$\sin(A + B)$ 用 $+$，$\sin(A - B)$ 用 $-$。
• Use cases.应用场合。 Exact values at $15^{\circ} = 45^{\circ} - 30^{\circ}$ and $75^{\circ} = 45^{\circ} + 30^{\circ}$; identity proofs that expand $\sin(\theta + \pi/2)$ etc.; AP Calc derivatives that need $\sin(x+h)$ expanded.$15^{\circ} = 45^{\circ} - 30^{\circ}$、$75^{\circ} = 45^{\circ} + 30^{\circ}$ 处的精确值；展开 $\sin(\theta + \pi/2)$ 等的恒等式证明；AP Calc 中需展开 $\sin(x+h)$ 的求导。

AB Math 30-1 indicator 6.7 names this use case verbatim: "determine, using the sum, difference and double-angle identities, the exact value of a trigonometric ratio." So worked-example 3 below answers a Math 30-1 indicator directly.AB Math 30-1 指标 6.7 原文点名："利用和差与二倍角恒等式确定三角比的精确值。" 故下方例题 3 直接对应 Math 30-1 一条指标。

Three formulas. The $\cos 2\theta$ formula has three equivalent forms.三条公式。$\cos 2\theta$ 有三种等价形式。 $$ \sin 2\theta = 2 \sin\theta \cos\theta $$ $$ \cos 2\theta = \cos^{2}\theta - \sin^{2}\theta = 2\cos^{2}\theta - 1 = 1 - 2\sin^{2}\theta $$ $$ \tan 2\theta = \frac{2 \tan\theta}{1 - \tan^{2}\theta} $$

• Where they come from.来源。 Set $A = B = \theta$ in the sum formulas of §3. The three $\cos 2\theta$ forms come from applying $\sin^{2}\theta + \cos^{2}\theta = 1$ to $\cos^{2}\theta - \sin^{2}\theta$ to eliminate either $\sin^{2}$ or $\cos^{2}$.在 §3 的和公式中取 $A = B = \theta$ 即可。$\cos 2\theta$ 的三种形式来自把 $\sin^{2}\theta + \cos^{2}\theta = 1$ 代入 $\cos^{2}\theta - \sin^{2}\theta$ 中消去 $\sin^{2}$ 或 $\cos^{2}$。
• Which $\cos 2\theta$ form to pick.$\cos 2\theta$ 形式的选择。 If the rest of the problem has $\cos\theta$, use $2\cos^{2}\theta - 1$. If it has $\sin\theta$, use $1 - 2\sin^{2}\theta$. Match the form to what is already in the expression.若题目其余部分有 $\cos\theta$，用 $2\cos^{2}\theta - 1$；若有 $\sin\theta$，用 $1 - 2\sin^{2}\theta$。让形式与现有式子匹配。
• Power-reduction.降次。 Rearrange to $\cos^{2}\theta = (1 + \cos 2\theta)/2$ and $\sin^{2}\theta = (1 - \cos 2\theta)/2$. AP Calc uses these for $\int \cos^{2} x \, dx$.整理得 $\cos^{2}\theta = (1 + \cos 2\theta)/2$ 与 $\sin^{2}\theta = (1 - \cos 2\theta)/2$。AP Calc 在 $\int \cos^{2} x \, dx$ 中用到。

AB Math 30-1 outcome 6 names double-angle (restricted to sine, cosine, tangent) as a required identity category. PC 12 Content elaboration lists "double angle" by name alongside Pythagorean, quotient, reciprocal, sum and difference.AB Math 30-1 目标 6 把二倍角（限于 sin、cos、tan）列为必学恒等式类别。PC 12 内容详释把"二倍角"与勾股、商、倒数、和差并列点名。

Square-root form. Sign chosen from the half-angle quadrant.根式形式。符号由半角所在象限决定。 $$ \sin\!\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos\theta}{2}} $$ $$ \cos\!\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos\theta}{2}} $$ $$ \tan\!\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}} \;=\; \frac{1 - \cos\theta}{\sin\theta} \;=\; \frac{\sin\theta}{1 + \cos\theta} $$

• Derivation.推导。 Start with $\cos 2\alpha = 1 - 2\sin^{2}\alpha$ from §4, set $\alpha = \theta/2$, isolate $\sin(\theta/2)$. Repeat with $\cos 2\alpha = 2\cos^{2}\alpha - 1$ to get $\cos(\theta/2)$.由 §4 的 $\cos 2\alpha = 1 - 2\sin^{2}\alpha$，取 $\alpha = \theta/2$ 分离 $\sin(\theta/2)$；再由 $\cos 2\alpha = 2\cos^{2}\alpha - 1$ 得 $\cos(\theta/2)$。
• Sign choice.符号选择。 The $\pm$ is decided by the quadrant of $\theta/2$, not the quadrant of $\theta$. If $\theta \in (0, 2\pi)$, then $\theta/2 \in (0, \pi)$, which can land in Q I or Q II.$\pm$ 由 $\theta/2$（而非 $\theta$）所在象限决定。若 $\theta \in (0, 2\pi)$，则 $\theta/2 \in (0, \pi)$，可能落在 Q I 或 Q II。
• Tangent shortcut.正切捷径。 The two non-square-root forms of $\tan(\theta/2)$ never need a $\pm$: their sign comes out automatically from $\sin\theta$ and $1 \pm \cos\theta$.$\tan(\theta/2)$ 的两种非根式形式不需 $\pm$：符号由 $\sin\theta$ 与 $1 \pm \cos\theta$ 自动给出。

Use case. Exact values at $\theta/2 = 22.5^{\circ}$ (from $\theta = 45^{\circ}$), $15^{\circ}$ (from $\theta = 30^{\circ}$), $75^{\circ}$ (from $\theta = 150^{\circ}$). AP Calc integration uses $\sin^{2}\theta = (1 - \cos 2\theta)/2$, which is the half-angle formula rewritten.应用场合。 $\theta/2 = 22.5^{\circ}$（由 $\theta = 45^{\circ}$）、$15^{\circ}$（由 $\theta = 30^{\circ}$）、$75^{\circ}$（由 $\theta = 150^{\circ}$）处的精确值。AP Calc 积分中用到 $\sin^{2}\theta = (1 - \cos 2\theta)/2$ , 即半角公式的另一种写法。

Pick one side. Reduce until it equals the other.选定一侧，化简至与另一侧相等。

A proof is a chain of equalities running from one side of the identity to the other, with every step justified by a known identity. Work on the more complex side. Never operate on both sides simultaneously , that risks assuming what you are trying to prove (AB Math 30-1 indicator 6.3 names this trap).证明是一串从一侧到另一侧的等式链，每一步用一个已知恒等式作依据。从较复杂的一侧入手。切勿同时对两侧操作 , 这等于先假设了你要证明的结论（AB Math 30-1 指标 6.3 明确点名此陷阱）。

1. Convert to $\sin$ and $\cos$.转为 $\sin$、$\cos$。 Rewrite every $\tan, \sec, \csc, \cot$ as a $\sin/\cos$ expression. Cancel and simplify. Handles roughly 60% of identities in this unit.把每个 $\tan, \sec, \csc, \cot$ 改写为 $\sin/\cos$；然后约分化简。可解本单元约 60% 的恒等式。
2. Use a Pythagorean identity.用勾股恒等式。 If you see $\sin^{2} + \cos^{2}$, replace with $1$. If you see $\sec^{2} - \tan^{2}$, replace with $1$. If you see $1 - \cos^{2}$, replace with $\sin^{2}$ (or vice versa).见 $\sin^{2} + \cos^{2}$ 替为 $1$；见 $\sec^{2} - \tan^{2}$ 替为 $1$；见 $1 - \cos^{2}$ 替为 $\sin^{2}$（或反之）。
3. Factor.因式分解。 If you see a difference of squares ($1 - \sin^{2}\theta = (1 - \sin\theta)(1 + \sin\theta)$), factor it. If you see $\sin\theta + \sin\theta \cos\theta$, factor out $\sin\theta$.见平方差（$1 - \sin^{2}\theta = (1 - \sin\theta)(1 + \sin\theta)$）就分解；见 $\sin\theta + \sin\theta \cos\theta$ 就提 $\sin\theta$。
4. Multiply by a conjugate.乘以共轭。 If you see $1 / (1 + \cos\theta)$, multiply numerator and denominator by $(1 - \cos\theta)$. The denominator becomes $1 - \cos^{2}\theta = \sin^{2}\theta$ , a quintessential identity move.见 $1 / (1 + \cos\theta)$ 时分子分母同乘 $(1 - \cos\theta)$。分母变为 $1 - \cos^{2}\theta = \sin^{2}\theta$ , 经典恒等式技巧。

AB Math 30-1 indicators 6.1, 6.3, 6.5, 6.6 together describe this workflow: identify the distinction between identity and equation (6.1), recognise that numerical verification is insufficient (6.3), state non-permissible values (6.5), and prove the identity algebraically (6.6).AB Math 30-1 指标 6.1、6.3、6.5、6.6 合起来即描述此流程：辨别恒等式与方程的区别（6.1）；认识到数值核验不充分（6.3）；写出非许可值（6.5）；代数证明（6.6）。

Three patterns. The interval rule decides what counts as a final answer.三种模式。区间规则决定最终答案的形式。

1. Linear in $\sin x$ (or $\cos x$).关于 $\sin x$ 的一次方程。 Isolate $\sin x = k$, then take $\arcsin k$ for one solution and $\pi - \arcsin k$ for the second on $[0, 2\pi)$ (or $180^{\circ} - \arcsin k$ in degree mode). Add $+ 2k\pi$ for the general solution.分离 $\sin x = k$；在 $[0, 2\pi)$ 上一解 $\arcsin k$、另一解 $\pi - \arcsin k$（角度模式：$180^{\circ} - \arcsin k$）。再 $+ 2k\pi$ 得通解。
2. Quadratic in $\sin x$ (or $\cos x$).关于 $\sin x$ 的二次方程。 Let $u = \sin x$, solve the quadratic in $u$, get $u = u_{1}, u_{2}$. For each $u_{i}$ in $[-1, 1]$, find $x$ as in the linear case. Discard any $|u_{i}| > 1$.设 $u = \sin x$ 求二次方程得 $u = u_{1}, u_{2}$；对 $[-1, 1]$ 范围内的 $u_{i}$ 按一次方程求 $x$；舍弃 $|u_{i}| > 1$ 的根。
3. Identity-required.需用恒等式。 If the equation mixes $\sin 2x$ with $\sin x$, or $\cos 2x$ with $\cos x$, use a double-angle identity to reduce to a single argument. Then factor or treat as quadratic.若方程同时含 $\sin 2x$ 与 $\sin x$（或 $\cos 2x$ 与 $\cos x$），用二倍角恒等式化为单一参数；再因式分解或当二次方程处理。

General solution vs interval restricted. Without an interval, give $x = \theta_{0} + 2k\pi$ (and $x = \pi - \theta_{0} + 2k\pi$ for sine, $x = -\theta_{0} + 2k\pi$ for cosine). With an interval like $[0, 2\pi)$, list every solution in that interval explicitly. AB Math 30-1 indicator 5.4 connects the general solution to the zeros of the trig function; indicator 5.5 names "determine the general solution" verbatim.通解与限定区间。 无区间时给出 $x = \theta_{0} + 2k\pi$（正弦还需 $x = \pi - \theta_{0} + 2k\pi$；余弦需 $x = -\theta_{0} + 2k\pi$）。给定区间如 $[0, 2\pi)$ 时，把区间内每个解显式列出。AB Math 30-1 指标 5.4 把通解与三角函数的零点联系；指标 5.5 原文点名"求通解"。

BC PC 12 Content elaboration covers this section in one line: "solving first- and second-degree equations (over restricted domains and all real numbers)." The "and all real numbers" half is the general-solution requirement.BC PC 12 内容详释一行覆盖本节："求解一次与二次三角方程（在限定域与全体实数上）。" "全体实数"对应通解要求。