5.2 Trigonometric Identities and Reduction Formulas

5.2 Trigonometric Identities and Reduction Formulas

Simplifying trigonometric expressions and evaluating exact values under constrained conditions are essential skills tested heavily in the Gaokao. This section explores fundamental same-angle identities, systematic reduction formulas (诱导公式), and algebraic transformation techniques tailored for competitive examination problems.

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1. Fundamental Same-Angle Identities (同角三角函数基本关系)

For any angle $\alpha$ where the relevant functions are defined, the coordinates $(x, y) = (\cos\alpha, \sin\alpha)$ of a point on the unit circle $x^2 + y^2 = 1$ satisfy two core identities:

1. Pythagorean Identity (平方关系)

$\sin^2\alpha + \cos^2\alpha = 1$

• Derived Forms: $\sin^2\alpha = 1 - \cos^2\alpha, \quad \cos^2\alpha = 1 - \sin^2\alpha$ $\sin\alpha = \pm\sqrt{1 - \cos^2\alpha}, \quad \cos\alpha = \pm\sqrt{1 - \sin^2\alpha}$ (Note: The choice of $\pm$ sign is strictly dictated by the quadrant in which $\alpha$ lies.)

2. Quotient Identity (商数关系)

$\tan\alpha = \frac{\sin\alpha}{\cos\alpha} \quad \left(\alpha \neq k\pi + \frac{\pi}{2}, k \in \mathbb{Z}\right)$

• Secondary Identity: Dividing $\sin^2\alpha + \cos^2\alpha = 1$ by $\cos^2\alpha$ gives: $1 + \tan^2\alpha = \frac{1}{\cos^2\alpha} = \sec^2\alpha \quad \left(\cos\alpha \neq 0\right)$

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2. Reduction Formulas (诱导公式)

Reduction formulas transform trigonometric functions of arbitrary angles into functions of acute angles $\alpha$. All reduction formulas involving integer multiples of $\frac{\pi}{2}$ (i.e., $k \cdot \frac{\pi}{2} \pm \alpha$) are governed by a single universal mnemonic rule.

Universal Mnemonic Rule: "奇变偶不变，符号看象限"

1. "奇变偶不变" (Odd changes, Even stays): Look at the integer coefficient $k$ of $\frac{\pi}{2}$ in $k \cdot \frac{\pi}{2} \pm \alpha$:
   • If $k$ is even ($k = 2m$, e.g., $\pi, 2\pi$), the function name remains unchanged ($\sin \to \sin$, $\cos \to \cos$, $\tan \to \tan$).
   • If $k$ is odd ($k = 2m+1$, e.g., $\frac{\pi}{2}, \frac{3\pi}{2}$), the function name changes to its co-function ($\sin \leftrightarrow \cos$).
2. "符号看象限" (Sign determined by quadrant): Assume $\alpha$ is an acute angle (Quadrant I). Determine which quadrant the composite angle $k \cdot \frac{\pi}{2} \pm \alpha$ falls into. The algebraic sign ($+$ or $-$) of the final result equals the sign of the original function in that specific quadrant.

Summary Table of Core Reduction Formulas

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3. High-Frequency Exam Techniques (高考核心技巧)

Technique 1: Evaluation of Homogeneous Expressions (齐次式求值)

When given $\tan\alpha = m$, expressions that are homogeneous polynomials in $\sin\alpha$ and $\cos\alpha$ can be evaluated without finding $\sin\alpha$ or $\cos\alpha$ individually.

1. Linear Homogeneous Fraction: $E = \frac{a\sin\alpha + b\cos\alpha}{c\sin\alpha + d\cos\alpha} = \frac{a\frac{\sin\alpha}{\cos\alpha} + b}{c\frac{\sin\alpha}{\cos\alpha} + d} = \frac{a\tan\alpha + b}{c\tan\alpha + d}$
2. Quadratic Homogeneous Expression: $F = a\sin^2\alpha + b\sin\alpha\cos\alpha + c\cos^2\alpha$ Multiply and divide by $1 = \sin^2\alpha + \cos^2\alpha$: $F = \frac{a\sin^2\alpha + b\sin\alpha\cos\alpha + c\cos^2\alpha}{\sin^2\alpha + \cos^2\alpha} = \frac{a\tan^2\alpha + b\tan\alpha + c}{\tan^2\alpha + 1}$

Technique 2: Symmetric Relations of $\sin\alpha$ and $\cos\alpha$ (对称式技巧)

Let $t = \sin\alpha + \cos\alpha$. Squaring both sides yields: $t^2 = (\sin\alpha + \cos\alpha)^2 = \sin^2\alpha + 2\sin\alpha\cos\alpha + \cos^2\alpha = 1 + 2\sin\alpha\cos\alpha$ $\implies \sin\alpha\cos\alpha = \frac{t^2 - 1}{2}$

• Domain Constraint on $t$: $t = \sin\alpha + \cos\alpha = \sqrt{2}\sin\left(\alpha + \frac{\pi}{4}\right) \implies t \in [-\sqrt{2}, \sqrt{2}]$
• Difference Identity: $(\sin\alpha - \cos\alpha)^2 = 1 - 2\sin\alpha\cos\alpha = 1 - (t^2 - 1) = 2 - t^2$ $\implies \sin\alpha - \cos\alpha = \pm\sqrt{2 - t^2}$
• Cubic Symmetric Sum: $\sin^3\alpha + \cos^3\alpha = (\sin\alpha + \cos\alpha)(1 - \sin\alpha\cos\alpha) = t \left(1 - \frac{t^2 - 1}{2}\right) = \frac{t(3 - t^2)}{2}$

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4. Worked Gaokao Exam Examples (高考典型例题精讲)

Worked Example 1: Homogeneous Expression Evaluation

Problem: Given $\tan\alpha = 2$, calculate the values of the following two expressions:

1. $E = \frac{3\sin\alpha - 5\cos\alpha}{2\sin\alpha + \cos\alpha}$
2. $F = 4\sin^2\alpha - 3\sin\alpha\cos\alpha + 2\cos^2\alpha$

Solution:

1. For Expression $E$: Divide both numerator and denominator by $\cos\alpha$ (since $\tan\alpha = 2 \implies \cos\alpha \neq 0$): $E = \frac{3\frac{\sin\alpha}{\cos\alpha} - 5}{2\frac{\sin\alpha}{\cos\alpha} + 1} = \frac{3\tan\alpha - 5}{2\tan\alpha + 1} = \frac{3(2) - 5}{2(2) + 1} = \frac{6 - 5}{4 + 1} = \frac{1}{5}$
2. For Expression $F$: Rewrite $F$ as a fraction over $\sin^2\alpha + \cos^2\alpha = 1$ and divide by $\cos^2\alpha$: $F = \frac{4\sin^2\alpha - 3\sin\alpha\cos\alpha + 2\cos^2\alpha}{\sin^2\alpha + \cos^2\alpha} = \frac{4\tan^2\alpha - 3\tan\alpha + 2}{\tan^2\alpha + 1}$ Substitute $\tan\alpha = 2$: $F = \frac{4(2^2) - 3(2) + 2}{2^2 + 1} = \frac{16 - 6 + 2}{4 + 1} = \frac{12}{5}$.

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Worked Example 2: Reduction Formula Simplification

Problem: Simplify the following trigonometric expression: $A = \frac{\sin(\pi - \alpha) \cdot \cos(\pi + \alpha) \cdot \tan(-\alpha)}{\cos\left(\frac{\pi}{2} + \alpha\right) \cdot \sin\left(\frac{3\pi}{2} - \alpha\right)}$

Solution:

1. Apply reduction formulas term by term:
   • $\sin(\pi - \alpha) = \sin\alpha$ (Quadrant II, sine is positive)
   • $\cos(\pi + \alpha) = -\cos\alpha$ (Quadrant III, cosine is negative)
   • $\tan(-\alpha) = -\tan\alpha$ (Quadrant IV, tangent is negative)
   • $\cos\left(\frac{\pi}{2} + \alpha\right) = -\sin\alpha$ (Odd multiple of $\frac{\pi}{2}$, Quadrant II, cosine is negative)
   • $\sin\left(\frac{3\pi}{2} - \alpha\right) = -\cos\alpha$ (Odd multiple of $\frac{\pi}{2}$, Quadrant III, sine is negative)
2. Substitute into numerator and denominator:
   • Numerator: $(\sin\alpha) \cdot (-\cos\alpha) \cdot (-\tan\alpha) = \sin\alpha \cos\alpha \tan\alpha = \sin^2\alpha$
   • Denominator: $(-\sin\alpha) \cdot (-\cos\alpha) = \sin\alpha \cos\alpha$
3. Simplify the quotient: $A = \frac{\sin^2\alpha}{\sin\alpha \cos\alpha} = \frac{\sin\alpha}{\cos\alpha} = \tan\alpha$