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

Secant, Cosecant, and Cotangent

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.

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

Cotangent

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

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

Graphs

• 
  Graph of sec x
• 
  Graph of cosec x
• 
  Graph of cot 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 $0<x<2\pi$

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

Identities

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 $1+\cot x\equiv \mathrm{cosec}x$.

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:

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

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.

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

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 $0<\theta <2\pi$

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

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

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

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

Notes