3.8 More Trigonometric Identities and EquationsHL ONLY

1. Reciprocal Trigonometric Functions

The HL syllabus introduces three reciprocal trigonometric functions.

Secant: $\sec\theta = \dfrac{1}{\cos\theta}$
Cosecant: $\csc\theta = \dfrac{1}{\sin\theta}$
Cotangent: $\cot\theta = \dfrac{1}{\tan\theta} = \dfrac{\cos\theta}{\sin\theta}$

Drawing a horizontal tangent line at $(0, 1)$ on the unit circle represents the cotangent. Extending the radius at angle $\theta$ to intersect this line gives a horizontal displacement equal to $\cot\theta$.

EXAMPLE 1 (Reciprocal Evaluation)

Given $\sin\theta = \dfrac{5}{13}$ and $\theta$ is an obtuse angle, evaluate $\sec\theta$ and $\cot\theta$.

Using the Pythagorean identity calculates the adjacent side:
$$\begin{aligned} \cos^2\theta &= 1 - \sin^2\theta \\ \cos^2\theta &= 1 - \left(\dfrac{5}{13}\right)^2 = \dfrac{144}{169} \\ \cos\theta &= -\dfrac{12}{13} \quad \text{(since $\theta$ is obtuse in Quadrant II)} \end{aligned}$$

Applying the definitions of the reciprocal functions yields:
$\sec\theta = \dfrac{1}{\cos\theta} \implies \mathbf{\sec\theta = -\dfrac{13}{12}}$
$\cot\theta = \dfrac{\cos\theta}{\sin\theta} \implies \mathbf{\cot\theta = -\dfrac{12}{5}}$

2. Advanced Pythagorean and Compound Identities

Rearranging the Pythagorean identity $\cos^2\theta + \sin^2\theta = 1$ derives the reciprocal identities:

Dividing by $\cos^2\theta$ gives: $\tan^2\theta + 1 = \sec^2\theta$.
Dividing by $\sin^2\theta$ gives: $\cot^2\theta + 1 = \csc^2\theta$.

Compound angle additions and subtractions require these expansions:

$\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$
$\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$
$\tan(A \pm B) = \dfrac{\tan A \pm \tan B}{1 \mp \tan A\tan B}$

Setting $A$ and $B$ equal produces the Double-Angle Identities:

$\sin(2A) = 2\sin A\cos A$
$\cos(2A) = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$
$\tan(2A) = \dfrac{2\tan A}{1 - \tan^2 A}$

EXAMPLE 2 (Compound Angle Expansions)

Evaluate $\sin 75^\circ$ and $\tan 15^\circ$ exactly without a calculator.

Part A: Evaluating $\sin 75^\circ$. $$\begin{aligned} \sin 75^\circ &= \sin(45^\circ + 30^\circ) \\ &= \sin 45^\circ\cos 30^\circ + \cos 45^\circ\sin 30^\circ \\ &= \left(\dfrac{\sqrt{2}}{2}\right)\left(\dfrac{\sqrt{3}}{2}\right) + \left(\dfrac{\sqrt{2}}{2}\right)\left(\dfrac{1}{2}\right) \\ &= \mathbf{\dfrac{\sqrt{6} + \sqrt{2}}{4}} \end{aligned}$$

Part B: Evaluating $\tan 15^\circ$. $$\begin{aligned} \tan 15^\circ &= \tan(60^\circ - 45^\circ) \\ &= \dfrac{\tan 60^\circ - \tan 45^\circ}{1 + \tan 60^\circ\tan 45^\circ} \\ &= \dfrac{\sqrt{3} - 1}{1 + \sqrt{3}(1)} \\ &= \mathbf{\dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}} \quad \text{(Rationalized to $2 - \sqrt{3}$)} \end{aligned}$$

3. General Multiple Variable Equalities

Solving equations of the form $\sin A = \sin B$ requires specific periodic rules.

Mathematical Condition	Conclusive Relation (Radians)
If $\sin A = \sin B$	$A = B + 2k\pi$ $A = (\pi - B) + 2k\pi$
If $\cos A = \cos B$	$A = B + 2k\pi$ $A = -B + 2k\pi$
If $\tan A = \tan B$	$A = B + k\pi$

Note: These relationships convert to degrees by substituting $180^\circ$ for $\pi$.

EXAMPLE 3 (Sine Variable-to-Variable Solutions)

Solve $\sin 3x = \sin x$ for $0 \le x \le 2\pi$.

The general solutions are given by two equations:

$$\begin{aligned} 3x &= x + 2k\pi \implies 2x = 2k\pi \implies x = k\pi \\ 3x &= \pi - x + 2k\pi \implies 4x = \pi + 2k\pi \implies x = \dfrac{\pi + 2k\pi}{4} \end{aligned}$$

Finding values within the domain $[0, 2\pi]$ gives:
$\mathbf{x = 0, \dfrac{\pi}{4}, \dfrac{3\pi}{4}, \pi, \dfrac{5\pi}{4}, \dfrac{7\pi}{4}, 2\pi}$.

EXAMPLE 4 (Cosine Variable-to-Variable Solutions)

Solve $\cos 3x = \cos x$ for $0 \le x \le 2\pi$.

The general solutions are:

$$\begin{aligned} 3x &= x + 2k\pi \implies 2x = 2k\pi \implies x = k\pi \\ 3x &= -x + 2k\pi \implies 4x = 2k\pi \implies x = \dfrac{k\pi}{2} \end{aligned}$$

Testing values of $k$ to find solutions within the domain gives:
$\mathbf{x = 0, \dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2}, 2\pi}$.

EXAMPLE 5 (Tangent Variable-to-Variable Solutions)

Solve $\tan 2x = \tan x$ for $0 \le x \le 2\pi$.

Using the tangent period rule gives:

$$\begin{aligned} 2x &= x + k\pi \implies x = k\pi \end{aligned}$$

Finding values within the domain gives:
$\mathbf{x = 0, \pi, 2\pi}$.

4. Converting Mismatched Trigonometric Functions

To solve equations with different trigonometric functions, like $\cos A = \sin B$, convert them to the same function using complementary identities:

$\sin x \equiv \cos\left(\dfrac{\pi}{2} - x\right) \quad \text{and} \quad \cos x \equiv \sin\left(\dfrac{\pi}{2} - x\right)$

To manage negative signs, use the odd/even parity properties:
$\sin(-x) = -\sin x$ (Odd function), $\tan(-x) = -\tan x$ (Odd function), but strictly $\cos(\pi - x) = -\cos x$ (adjusting boundary properties).

EXAMPLE 6 (Mixed Functions)

Solve $\cos 3x = \sin x$ for $0 \le x \le 2\pi$.

Convert to a uniform function using a complementary identity:
$\cos 3x = \cos\left(\dfrac{\pi}{2} - x\right)$

Set up the general equations:

$$\begin{aligned} 3x &= \left(\dfrac{\pi}{2} - x\right) + 2k\pi \implies 4x = \dfrac{\pi}{2} + 2k\pi \implies x = \dfrac{\pi}{8} + \dfrac{k\pi}{2} \\ 3x &= -\left(\dfrac{\pi}{2} - x\right) + 2k\pi \implies 2x = -\dfrac{\pi}{2} + 2k\pi \implies x = -\dfrac{\pi}{4} + k\pi \end{aligned}$$

Substitute integer values for $k$ to find solutions within the domain.

EXAMPLE 7 (Mixed Signs)

Solve $\sin 3x = -\sin x$ for $0 \le x \le 2\pi$.

Use the odd function property to absorb the negative sign:
$\sin 3x = \sin(-x)$

Set up the general equations:

$$\begin{aligned} 3x &= -x + 2k\pi \implies 4x = 2k\pi \implies x = \dfrac{k\pi}{2} \\ 3x &= \pi - (-x) + 2k\pi \implies 2x = \pi + 2k\pi \implies x = \dfrac{\pi}{2} + k\pi \end{aligned}$$

Substitute values for $k$ to find solutions within the bounds.

EXAMPLE 8 (Resolving Reciprocal Equations)

Solve for $\sec x = 2$.

Solution: The equation translates to $\cos x = \dfrac{1}{2}$.

The general solution in radians is:
$\mathbf{x = \pm\dfrac{\pi}{3} + 2k\pi}$

The general solution in degrees is:
$\mathbf{x = \pm 60^\circ + 360^\circ k}$