M203 20260207 Trigonometric Identities

📝 Homework & Extensions

Suppose that $\sec x + \tan x = a$, find $\csc x + \cot x$

Weierstrass Substitution Core

By setting $t = \tan(x/2)$, all trigonometric functions can be expressed as rational functions of $t$:

$$\sin x = \frac{2t}{1+t^2}, \quad \cos x = \frac{1-t^2}{1+t^2}, \quad \tan x = \frac{2t}{1-t^2}$$

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Application to the Question

1. Given: $\sec x + \tan x = a$

Using the substitution:

$$\frac{1+t^2}{1-t^2} + \frac{2t}{1-t^2} = \frac{t^2+2t+1}{1-t^2} = \frac{(1+t)^2}{(1+t)(1-t)} = \frac{1+t}{1-t}$$

Using the tangent addition formula $\tan(x/2 + \pi/4) = \dfrac{\tan(x/2) + \tan(\pi/4)}{1 - \tan(x/2)\tan(\pi/4)} = \dfrac{t+1}{1-t}$, we find:

$$a = \frac{1+t}{1-t}$$

2. Target: $\csc x + \cot x$

Using the substitution:

$$\frac{1+t^2}{2t} + \frac{1-t^2}{2t} = \frac{2}{2t} = \frac{1}{t}$$

Alternatively, via the half-angle identity: $\cot(x/2) = \frac{1+\cos x}{\sin x} = \csc x + \cot x$.

3. Final Solution

Solve for $t$ in terms of $a$ from Step 1:

$$a(1-t) = 1+t \implies a - at = 1+t \implies a-1 = t(a+1) \implies t = \frac{a-1}{a+1}$$

Since the target is $\frac{1}{t}$:

$$\csc x + \cot x = \frac{1}{t} = \mathbf{\frac{a+1}{a-1}}$$

The Auxiliary Angle Formula is a powerful tool used to combine a linear combination of sine and cosine into a single trigonometric function. This is essential for solving equations, finding maximum/minimum values, and simplifying complex expressions in competitions like the AMC.

The general formula is:

$$a \sin x + b \cos x = \sqrt{a^2 + b^2} \sin(x + \phi)$$

Where the phase shift $\phi$ is determined by:

$$\cos \phi = \frac{a}{\sqrt{a^2 + b^2}} \quad \text{and} \quad \sin \phi = \frac{b}{\sqrt{a^2 + b^2}} \implies \tan \phi = \frac{b}{a}$$

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Example 1: Finding the Maximum Value

Problem: Find the maximum value of $f(x) = 3 \sin x + 4 \cos x$.

Solution:

1. Identify $a=3$ and $b=4$.

2. Calculate the amplitude: $\sqrt{3^2 + 4^2} = \sqrt{25} = 5$.

3. The expression becomes: $5 \sin(x + \phi)$, where $\tan \phi = \frac{4}{3}$.

4. Since the range of a sine function is $[-1, 1]$, the range of $5 \sin(x + \phi)$ is $[-5, 5]$.

Result: The maximum value is 5.

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Example 2: Solving an Equation with Special Angles

Problem: Solve $\sin x - \sqrt{3} \cos x = 1$ for $0 \le x < 2\pi$.

Solution:

1. $a=1, b=-\sqrt{3}$. Amplitude: $\sqrt{1^2 + (-\sqrt{3})^2} = 2$.

2. Factor out the 2: $2 \left( \frac{1}{2} \sin x - \frac{\sqrt{3}}{2} \cos x \right) = 1$.

3. Recognize $\cos(60^\circ) = \frac{1}{2}$ and $\sin(60^\circ) = \frac{\sqrt{3}}{2}$.

4. Apply the sine difference formula: $2 \sin(x - 60^\circ) = 1$.

5. $\sin(x - 60^\circ) = \frac{1}{2}$.

• $x - 60^\circ = 30^\circ \implies x = 90^\circ$ ($\frac{\pi}{2}$)

• $x - 60^\circ = 150^\circ \implies x = 210^\circ$ ($\frac{7\pi}{6}$)

Result: $x = 90^\circ, 210^\circ$.

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Example 3: The $\sqrt{2}$ Shortcut (Very Common in AMC)

Problem: Simplify $\sin x + \cos x$.

Solution:

When $a=1$ and $b=1$, the amplitude is $\sqrt{2}$ and the angle $\phi$ is $45^\circ$ (or $\frac{\pi}{4}$).

$$\sin x + \cos x = \sqrt{2} \sin(x + 45^\circ)$$

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Summary Table for Quick Reference

To master the Triple Angle Formulas for competitions like the AMC 10/12, you need both the standard forms and the "hidden" product forms that frequently appear in simplification problems.

1. The Standard Formulas

These are the foundational identities. A common mnemonic to remember the coefficients is "43" for Cosine and "34" for Sine.

• Sine Triple Angle:

$$\sin 3x = 3\sin x - 4\sin^3 x$$

• Cosine Triple Angle:

$$\cos 3x = 4\cos^3 x - 3\cos x$$

• Tangent Triple Angle:

$$\tan 3x = \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x}$$

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2. The "Hidden" Product Trick (High-Level)

This is the most powerful version for competition math. It allows you to convert the triple angle into a product of three terms, which is perfect for canceling terms in fractions.

$$\sin 3x = 4\sin x \sin(60^\circ - x) \sin(60^\circ + x)$$

$$\cos 3x = 4\cos x \cos(60^\circ - x) \cos(60^\circ + x)$$

$$\tan 3x = \tan x \tan(60^\circ - x) \tan(60^\circ + x)$$

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3. The Ratio Trick

As we used in a previous problem, the ratio of a triple angle to its single angle is often just a shifted version of $\cos 2x$. This is a huge time-saver!

• Sine Ratio:

$$\frac{\sin 3x}{\sin x} = 3 - 4\sin^2 x = 2\cos 2x + 1$$

• Cosine Ratio:

$$\frac{\cos 3x}{\cos x} = 4\cos^2 x - 3 = 2\cos 2x - 1$$

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4. Practice Application

Problem: Simplify $4\cos^3 20^\circ - 3\cos 20^\circ$.

Solution:

Recognize the pattern $4\cos^3 \theta - 3\cos \theta$. This is exactly the formula for $\cos 3\theta$.

• Set $\theta = 20^\circ$.

• Result $= \cos(3 \cdot 20^\circ) = \cos 60^\circ = \frac{1}{2}$.

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Summary Table