5.3 Addition, Subtraction, and Double-Angle Formulas

5.3 Addition, Subtraction, and Double-Angle Formulas

Compound angle formulas represent the foundational toolkit for algebraic manipulations in upper secondary trigonometry. In the Gaokao exam, these formulas serve as indispensable bridges to reduce complex trigonometric functions into standard forms $y = A\sin(\omega x + \phi) + B$ and solve non-standard numerical calculations.

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1. Sum and Difference Formulas (两角和与差的公式)

Vector Derivation of $\cos(\alpha - \beta)$

Consider two unit vectors $\vec{u} = (\cos\alpha, \sin\alpha)$ and $\vec{v} = (\cos\beta, \sin\beta)$ whose initial points are at the origin. The angle between $\vec{u}$ and $\vec{v}$ is $|\alpha - \beta|$.

• Dot Product Definition: $\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos(\alpha - \beta) = (1)(1)\cos(\alpha - \beta) = \cos(\alpha - \beta)$
• Coordinate Dot Product: $\vec{u} \cdot \vec{v} = \cos\alpha \cos\beta + \sin\alpha \sin\beta$ Equating both forms yields the fundamental difference formula for cosine: $\cos(\alpha - \beta) = \cos\alpha \cos\beta + \sin\alpha \sin\beta$

Complete Set of Sum and Difference Formulas

Replacing $\beta$ with $-\beta$ and using co-function identities yields the complete system:

1. Cosine Sum & Difference: $\cos(\alpha + \beta) = \cos\alpha \cos\beta - \sin\alpha \sin\beta$ $\cos(\alpha - \beta) = \cos\alpha \cos\beta + \sin\alpha \sin\beta$
2. Sine Sum & Difference: $\sin(\alpha + \beta) = \sin\alpha \cos\beta + \cos\alpha \sin\beta$ $\sin(\alpha - \beta) = \sin\alpha \cos\beta - \cos\alpha \sin\beta$
3. Tangent Sum & Difference: $\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha \tan\beta}$ $\tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha \tan\beta}$
   • Useful Deformation for Gaokao: $\tan\alpha + \tan\beta = \tan(\alpha + \beta)(1 - \tan\alpha \tan\beta)$

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2. Double-Angle, Power-Reduction, and Half-Angle Formulas (倍角、降幂与半角公式)

1. Double-Angle Formulas (二倍角公式)

Setting $\beta = \alpha$ in the sum formulas yields:

• Sine Double-Angle: $\sin 2\alpha = 2\sin\alpha \cos\alpha$
• Cosine Double-Angle (Three Equivalent Forms): $\cos 2\alpha = \cos^2\alpha - \sin^2\alpha = 2\cos^2\alpha - 1 = 1 - 2\sin^2\alpha$
• Tangent Double-Angle: $\tan 2\alpha = \frac{2\tan\alpha}{1 - \tan^2\alpha}$

2. Power-Reduction Formulas (降幂公式)

Rearranging the cosine double-angle formula lowers quadratic terms to linear terms (essential for graphing and integration): $\sin^2\alpha = \frac{1 - \cos 2\alpha}{2}$ $\cos^2\alpha = \frac{1 + \cos 2\alpha}{2}$ $\sin\alpha \cos\alpha = \frac{1}{2} \sin 2\alpha$

3. Half-Angle Formulas (半角公式)

Replacing $\alpha$ with $\frac{\alpha}{2}$ in the power-reduction identities: $\sin\frac{\alpha}{2} = \pm\sqrt{\frac{1 - \cos\alpha}{2}}, \quad \cos\frac{\alpha}{2} = \pm\sqrt{\frac{1 + \cos\alpha}{2}}$ $\tan\frac{\alpha}{2} = \frac{1 - \cos\alpha}{\sin\alpha} = \frac{\sin\alpha}{1 + \cos\alpha}$

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3. The Auxiliary Angle Formula (辅助角公式 / 提辅角法)

One of the most frequently examined techniques in Gaokao trigonometry is converting linear combinations of sine and cosine of the same frequency into a single sinusoidal function.

Theorem

For real constants $a, b$ (not both zero): $a\sin\theta + b\cos\theta = \sqrt{a^2 + b^2} \sin(\theta + \phi)$ where the auxiliary angle $\phi$ satisfies: $\cos\phi = \frac{a}{\sqrt{a^2 + b^2}}, \quad \sin\phi = \frac{b}{\sqrt{a^2 + b^2}}, \quad \tan\phi = \frac{b}{a}$

Proof Sketch

Factor out $\sqrt{a^2 + b^2}$: $a\sin\theta + b\cos\theta = \sqrt{a^2 + b^2} \left( \frac{a}{\sqrt{a^2 + b^2}} \sin\theta + \frac{b}{\sqrt{a^2 + b^2}} \cos\theta \right)$ Since $\left(\frac{a}{\sqrt{a^2+b^2}}\right)^2 + \left(\frac{b}{\sqrt{a^2+b^2}}\right)^2 = 1$, there exists a unique angle $\phi \in (-\pi, \pi]$ such that $\cos\phi = \frac{a}{\sqrt{a^2+b^2}}$ and $\sin\phi = \frac{b}{\sqrt{a^2+b^2}}$. Applying the sine sum formula gives $\sqrt{a^2+b^2}(\sin\theta \cos\phi + \cos\theta \sin\phi) = \sqrt{a^2+b^2}\sin(\theta + \phi)$.

Special Common Values

1. $\sin\theta + \cos\theta = \sqrt{2} \sin\left(\theta + \frac{\pi}{4}\right)$
2. $\sqrt{3}\sin\theta + \cos\theta = 2 \sin\left(\theta + \frac{\pi}{6}\right)$
3. $\sin\theta - \sqrt{3}\cos\theta = 2 \sin\left(\theta - \frac{\pi}{3}\right)$

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

Worked Example 1: Function Reduction using Auxiliary Angle Formula

Problem: Given the function $f(x) = 2\sin x \cos x + 2\sqrt{3}\cos^2 x - \sqrt{3}$.

1. Reduce $f(x)$ into the form $A\sin(\omega x + \phi)$.
2. Find the maximum value of $f(x)$ and the values of $x \in [0, \pi]$ at which the maximum occurs.

Solution:

1. Apply double-angle and power-reduction formulas:
   • $2\sin x \cos x = \sin 2x$
   • $2\sqrt{3}\cos^2 x - \sqrt{3} = \sqrt{3}(2\cos^2 x - 1) = \sqrt{3}\cos 2x$ Substituting these yields: $f(x) = \sin 2x + \sqrt{3}\cos 2x$
2. Apply the auxiliary angle formula: Here $a = 1$ and $b = \sqrt{3}$. The amplitude is $\sqrt{1^2 + (\sqrt{3})^2} = 2$. $\cos\phi = \frac{1}{2}, \quad \sin\phi = \frac{\sqrt{3}}{2} \implies \phi = \frac{\pi}{3}$ $f(x) = 2 \sin\left(2x + \frac{\pi}{3}\right)$
3. Find the maximum value on $x \in [0, \pi]$: Since $x \in [0, \pi]$, the phase $2x + \frac{\pi}{3}$ ranges over: $2x + \frac{\pi}{3} \in \left[\frac{\pi}{3}, 2\pi + \frac{\pi}{3}\right] = \left[\frac{\pi}{3}, \frac{7\pi}{3}\right]$ The sine function achieves its maximum value of $1$ when the phase equals $\frac{\pi}{2}$ or $\frac{5\pi}{2}$.
   • Setting $2x + \frac{\pi}{3} = \frac{\pi}{2} \implies 2x = \frac{\pi}{6} \implies x = \frac{\pi}{12}$.
   • Maximum value is $f\left(\frac{\pi}{12}\right) = 2(1) = 2$.

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Worked Example 2: Compound Angle Parameter Calculation

Problem: Let $\alpha, \beta \in \left(0, \frac{\pi}{2}\right)$ be acute angles such that $\cos\alpha = \frac{4}{5}$ and $\cos(\alpha + \beta) = -\frac{3}{5}$. Find $\sin\beta$ and the exact value of $\beta$.

Solution:

1. Determine $\sin\alpha$: Since $\alpha \in \left(0, \frac{\pi}{2}\right)$, $\sin\alpha > 0$: $\sin\alpha = \sqrt{1 - \cos^2\alpha} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \frac{3}{5}$
2. Determine $\sin(\alpha + \beta)$: Since $\alpha, \beta \in \left(0, \frac{\pi}{2}\right)$, we have $\alpha + \beta \in (0, \pi)$. Given $\cos(\alpha + \beta) = -\frac{3}{5} < 0$, $\alpha + \beta$ must lie in Quadrant II, so $\sin(\alpha + \beta) > 0$: $\sin(\alpha + \beta) = \sqrt{1 - \cos^2(\alpha + \beta)} = \sqrt{1 - \left(-\frac{3}{5}\right)^2} = \frac{4}{5}$
3. Express $\beta = (\alpha + \beta) - \alpha$ and apply difference formula: $\cos\beta = \cos((\alpha + \beta) - \alpha) = \cos(\alpha + \beta)\cos\alpha + \sin(\alpha + \beta)\sin\alpha$ $\cos\beta = \left(-\frac{3}{5}\right)\left(\frac{4}{5}\right) + \left(\frac{4}{5}\right)\left(\frac{3}{5}\right) = -\frac{12}{25} + \frac{12}{25} = 0$
4. Conclusion: Since $\beta \in \left(0, \frac{\pi}{2}\right)$ and $\cos\beta = 0$, we have $\beta = \frac{\pi}{2}$ (or $\sin\beta = 1$).