A-level Mathematics/CIE/Pure Mathematics 2/Trigonometry

Secant
The secant of an angle is the reciprocal of its cosine.

$$\sec x = \frac{1}{\cos x}$$

Cosecant
The cosecant of an angle is the reciprocal of its sine.

$$\operatorname{cosec} x = \frac{1}{\sin x}$$

Cotangent
The cotangent of an angle is the reciprocal of its tangent.

$$\cot x = \frac{1}{\tan x}$$

Solving Equations with Secants, Cosecants, and Cotangents
Solving an equation with secants, cosecants, or cotangents is pretty much the same method as with any other trigonometric equation.

e.g. Solve $$-\sec x = 2 + \cos x$$ for $$0 < x < 2\pi$$

$$\begin{align} -\sec x &= 2 + \cos x \\ -1 &= 2\cos x + \cos^2 x \\ \cos^2 x + 2\cos x + 1 &= 0 \\ (\cos x + 1)^2 &= 0 \\ \cos x + 1 &= 0 \\ \cos x &= -1 \\ x &= \pi \end{align}$$

Cotangent identity
$$\tan x \equiv \frac{\sin x}{\cos x}$$ and $$\cot x \equiv \frac{1}{\tan x}$$, therefore $$\cot x \equiv \frac{\cos x}{\sin x}$$

Pythagorean-derived identities
The Pythagorean trigonometric identity states that $$\sin^2 x + \cos^2 x \equiv 1$$. We can divide both sides by $$\cos^2 x$$ to obtain another identity: $$\tan^2 x + 1 \equiv \sec^2 x$$. Alternatively, we can divide both sides by $$\sin^2 x$$ to obtain $$1 + \cot x \equiv \operatorname{cosec} x$$.

Addition Formulae
The addition formulae are used when we have a trigonometric function applied to a sum or difference, e.g. $$\sin(\theta+\frac{\pi}{6})$$.

For sine, cosine, and tangent, the addition formulae are:

$$\begin{align} \sin(A\pm B) &= \sin A\cos B \pm \cos A\sin B \\ \cos(A\pm B) &= \cos A\cos B \mp \sin A\sin B \\ \tan(A\pm B) &= \frac{\tan A \pm \tan B}{1\mp\tan A\tan B} \end{align}$$

Double Angle Formulae
The double angle formulae are a special case of the addition formulae, when both of the terms in the sum are equal.

$$\begin{align} \sin(2A) &= \sin(A+A) = \sin A\cos A + \sin A\cos A = 2\sin A\cos A \\ \cos(2A) &= \cos(A+A) = \cos A\cos A - \sin A\sin A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A \\ \tan(2A) &= \tan(A+A) = \frac{\tan A + \tan A}{1 - \tan A\tan A} = \frac{2\tan A}{1-\tan^2 A} \end{align}$$

Converting $$a\sin\theta + b\cos\theta$$ to $$R\sin(\theta\pm\alpha)$$ or $$R\cos(\theta\pm\alpha)$$
It is helpful when solving trigonometric equations to convert an expression into a single term. To do this, we can use the addition formulae.

e.g. Solve $$\sin\theta + \sqrt{3}\cos\theta = 1$$ for $$0 < \theta < 2\pi$$

$$\begin{align} \sin\theta + \sqrt{3}\cos\theta &= R\sin(\theta + \alpha) \\ \sin\theta + \sqrt{3}\cos\theta &= R\sin\theta\cos\alpha + R\cos\theta\sin\alpha \\ \text{Equate coefficients of }\sin\theta\to\quad R\cos\alpha &= 1 \\ \text{Equate coefficients of }\cos\theta\to\quad R\sin\alpha &= \sqrt{3} \\ \tan\alpha &= \frac{R\sin\alpha}{R\cos\alpha} = \frac{\sqrt{3}}{1} \\ \therefore\alpha &= \frac{\pi}{3} \\ R\cos(\frac{\pi}{3}) &= 1 \\ R &= \frac{1}{\tfrac{1}{2}} = 2 \\ 2\sin(\theta + \frac{\pi}{3}) &= 1 \\ \sin(\theta + \frac{\pi}{3}) &= \frac{1}{2} \\ \theta + \frac{\pi}{3} &= \left\{\frac{5\pi}{6}, \frac{13\pi}{6}\right\} \leftarrow \text{Be careful with the domain of }\theta + \frac{\pi}{3} \\ \theta &= \left\{\frac{3\pi}{6}, \frac{11\pi}{6}\right\} \end{align}$$

Using $$R\cos(\theta\pm\alpha)$$ is pretty similar.

e.g. Solve $$\sqrt{2}\sin\theta + \sqrt{2}\cos\theta = 1$$ for $$0 < \theta < 2\pi$$

$$\begin{align} \sqrt{2}\sin\theta + \sqrt{2}\cos\theta &= R\cos(\theta - \alpha) \\ \sqrt{2}\sin\theta + \sqrt{2}\cos\theta &= R\cos\theta\cos\alpha + R\sin\theta\sin\alpha \\ \text{Equate coefficients of}\sin\theta\to\quad R\sin\alpha &= \sqrt{2} \\ \text{Equate coefficients of}\cos\theta\to\quad R\cos\alpha &= \sqrt{2} \\ \tan\alpha &= \frac{R\sin\alpha}{R\cos\alpha} = \frac{\sqrt{2}}{\sqrt{2}} = 1 \\ \therefore\alpha &= \frac{\pi}{4} \\ R\sin\left(\frac{\pi}{4}\right) &= \sqrt{2} \\ R\left(\frac{1}{\sqrt{2}}\right) &= \sqrt{2} \\ R &= \sqrt{2}(\sqrt{2}) = 2 \\ 2\cos(\theta - \frac{\pi}{4}) &= 1 \\ \cos(\theta - \frac{\pi}{4}) &= \frac{1}{2} \\ \theta - \frac{\pi}{4} &= \left\{\frac{\pi}{3},\frac{5\pi}{3}\right\} \\ \theta &= \left\{\frac{7\pi}{12},\frac{23\pi}{12}\right\} \end{align}$$


 * Notes