Further Trigonometry — IB Math AA (HL)

1. Reciprocal Trigonometric Functions

The three reciprocal functions are defined from the primary trig functions and appear in many HL calculus and integration contexts.

Reciprocal Functions

$\sec x = \dfrac{1}{\cos x}$ $\csc x = \dfrac{1}{\sin x}$ $\cot x = \dfrac{1}{\tan x} = \dfrac{\cos x}{\sin x}$

Pythagorean identities:

$1 + \tan^2 x = \sec^2 x$ (from $\sin^2 x + \cos^2 x = 1$, dividing by $\cos^2 x$)

$1 + \cot^2 x = \csc^2 x$ (from $\sin^2 x + \cos^2 x = 1$, dividing by $\sin^2 x$)

2. Inverse Trigonometric Functions

Inverse Trig: Domains and Ranges

$y = \arcsin x$: domain $[-1, 1]$, range $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$

$y = \arccos x$: domain $[-1, 1]$, range $[0, \pi]$

$y = \arctan x$: domain $\mathbb{R}$, range $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$

The restricted domains ensure these are genuine functions (one-to-one). The ranges are called the principal values. Note: $\arcsin(\sin x) = x$ only when $x \in \left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.

3. Compound Angle Formulas

Compound Angle Identities

$\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$

$\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$

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

Double angle formulas:

$\sin 2A = 2\sin A \cos A$

$\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$

$\tan 2A = \dfrac{2\tan A}{1-\tan^2 A}$

Power-Reduction: From double angle: $\cos^2 A = \dfrac{1+\cos 2A}{2}$ and $\sin^2 A = \dfrac{1-\cos 2A}{2}$. These are essential for integrating $\sin^2 x$ and $\cos^2 x$.

4. The $R\sin(x + \varphi)$ Form

Any expression of the form $a\sin x + b\cos x$ can be written as $R\sin(x+\varphi)$ where:

$$R = \sqrt{a^2+b^2}, \quad \tan\varphi = \frac{b}{a}$$

This is useful for finding maximum/minimum values and solving equations.

Worked Example 1

Express $3\sin x + 4\cos x$ in the form $R\sin(x+\varphi)$

Expand $R\sin(x+\varphi) = R\sin x\cos\varphi + R\cos x\sin\varphi$. Comparing: $R\cos\varphi = 3$ and $R\sin\varphi = 4$.

Find $R$: $R = \sqrt{3^2+4^2} = \sqrt{25} = 5$

Find $\varphi$: $\tan\varphi = \dfrac{4}{3} \Rightarrow \varphi = \arctan\!\left(\dfrac{4}{3}\right) \approx 53.1° = 0.927$ rad

Result: $3\sin x + 4\cos x = 5\sin(x + 0.927)$. The maximum value is $5$ and minimum is $-5$.

Worked Example 2

Solve $2\sin^2 x + \sin x - 1 = 0$ for general solution

Factor: $(2\sin x - 1)(\sin x + 1) = 0$

Case 1: $\sin x = \dfrac{1}{2} \Rightarrow x = \dfrac{\pi}{6}$ or $x = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$ in $[0, 2\pi)$

Case 2: $\sin x = -1 \Rightarrow x = \dfrac{3\pi}{2}$ in $[0,2\pi)$

General solutions: $x = \dfrac{\pi}{6} + 2n\pi$, $x = \dfrac{5\pi}{6} + 2n\pi$, $x = \dfrac{3\pi}{2} + 2n\pi = -\dfrac{\pi}{2} + 2n\pi$, for $n \in \mathbb{Z}$.

Worked Example 3

Prove: $\dfrac{\cos 2\theta}{1+\sin 2\theta} = \dfrac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}$

LHS numerator: $\cos 2\theta = \cos^2\theta - \sin^2\theta = (\cos\theta+\sin\theta)(\cos\theta-\sin\theta)$

LHS denominator: $1 + \sin 2\theta = 1 + 2\sin\theta\cos\theta = \sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = (\sin\theta+\cos\theta)^2$

Combine: $\text{LHS} = \dfrac{(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)}{(\sin\theta+\cos\theta)^2} = \dfrac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta} = \text{RHS}$ $\square$

Practice Problems

Q1. Write $\sqrt{3}\cos x - \sin x$ in the form $R\cos(x+\alpha)$ and find its maximum value and the smallest positive $x$ where it occurs.

Show Solution

$R\cos(x+\alpha) = R\cos x\cos\alpha - R\sin x\sin\alpha$. So $R\cos\alpha = \sqrt{3}$ and $R\sin\alpha = 1$.

$R = \sqrt{3+1} = 2$, $\tan\alpha = \dfrac{1}{\sqrt{3}} \Rightarrow \alpha = \dfrac{\pi}{6}$.

Expression $= 2\cos\!\left(x+\dfrac{\pi}{6}\right)$. Maximum value is $2$, occurring when $x + \pi/6 = 0$, i.e., $x = -\pi/6$. Smallest positive: $x = 2\pi - \pi/6 = 11\pi/6$.

Q2. Prove the identity $\sec^2 x + \csc^2 x = \sec^2 x \csc^2 x$.

Show Solution

LHS $= \dfrac{1}{\cos^2 x} + \dfrac{1}{\sin^2 x} = \dfrac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} = \dfrac{1}{\sin^2 x \cos^2 x} = \dfrac{1}{\cos^2 x} \cdot \dfrac{1}{\sin^2 x} = \sec^2 x \csc^2 x$ $\square$

Q3. Find the general solution of $\cos 2x = \cos x$.

Show Solution

$2\cos^2 x - 1 = \cos x \Rightarrow 2\cos^2 x - \cos x - 1 = 0 \Rightarrow (2\cos x + 1)(\cos x - 1) = 0$.

$\cos x = 1 \Rightarrow x = 2n\pi$. $\cos x = -\dfrac{1}{2} \Rightarrow x = \pm\dfrac{2\pi}{3} + 2n\pi$. All $n \in \mathbb{Z}$.

5. Exam Tips

When solving trig equations, always use a CAST diagram or unit circle to find all solutions in the required interval.
For general solutions: $\sin\theta = k$ gives $\theta = \arcsin k + 2n\pi$ and $\theta = \pi - \arcsin k + 2n\pi$.
In $R\sin(x+\varphi)$ form, always verify the sign of $\varphi$ is consistent with both $R\cos\varphi$ and $R\sin\varphi$.
The identities $1+\tan^2\theta = \sec^2\theta$ and $1+\cot^2\theta = \csc^2\theta$ are the most-used reciprocal identities in integration.