Basic Identity

Basic Identities

6.4

Basic Identities

There are three basic identities as follows:


	\(\text{sin}^2 \ \theta + \text{cos}^2 \ \theta = 1\)

	\(1 + \text{tan}^2 \ \theta = \text{sec}^2 \ \theta \)

	\(1 + \text{cot}^2 \ \theta = \text{cosec}^2 \ \theta \)

All three basic identities can be derived by using a right-angled triangle \(ABC\) and all the trigonometric ratios which have been learnt


	\(\text{sin }A = \dfrac{a}{c}\)		\(\text{cosec }A = \dfrac{c}{a}\)

	\(\text{cos }A = \dfrac{b}{c}\)		\(\text{sec }A = \dfrac{c}{b}\)

	\(\text{tan }A = \dfrac{a}{b}\)		\(\text{cot }A = \dfrac{b}{a}\)

By using Pythagoras theorem, it is known that


	\(a^2+b^2 = c^2\)

Divide the two sides of the equation by \(a^2, \ b^2 \text{ and }c^2\), we get:

\(\div \ a^2\)

\(\div \ b^2\)

\(\div \ c^2\)


	\(\begin{aligned} &\dfrac{a^2}{a^2}+ \dfrac{b^2}{a^2} = \dfrac{c^2}{a^2}\\\\ &1 + \begin{pmatrix} \dfrac{b}{a} \end{pmatrix}^2 = \begin{pmatrix} \dfrac{c}{a} \end{pmatrix}^2\\\\ &1+ \text{cot}^2 \ A = \text{cosec}^2 \ A \end{aligned}\)


	\(\begin{aligned}& \dfrac{a^2}{b^2}+ \dfrac{b^2}{b^2} = \dfrac{c^2}{b^2}\\\\ &\begin{pmatrix} \dfrac{a}{b} \end{pmatrix}^2 + 1= \begin{pmatrix} \dfrac{c}{b} \end{pmatrix}^2\\\\ &1+ \text{tan}^2 \ A = \text{sec}^2 \ A \end{aligned}\)


	\(\begin{aligned} &\dfrac{a^2}{c^2}+ \dfrac{b^2}{c^2} = \dfrac{c^2}{c^2}\\\\ &\begin{pmatrix} \dfrac{a}{c} \end{pmatrix}^2+ \begin{pmatrix} \dfrac{b}{c} \end{pmatrix}^2 = 1\\\\ &\text{sin}^2 \ A+ \text{cos}^2 \ A = 1 \end{aligned}\)

These three basic trigonometric identities can be used to solve problems involving trigonometric ratios

Example

Without using a calculator, find the value of each of the following:

(a)	\(\text{sin}^2 (-430^{\circ}) + \text{cos}^2(-430^{\circ}) \)

(b)	\(\text{tan}^2 \begin{pmatrix} \dfrac{\pi}{3} \end{pmatrix} - \text{sec}^2\begin{pmatrix} \dfrac{\pi}{3} \end{pmatrix} \)


Solution:

(a)	\(\text{sin}^2 \ A + \text{cos}^2 \ A=1\\ \text{sin}^2 (-430^{\circ}) + \text{cos}^2(-430^{\circ}) =1\)

(b)	\(\begin{aligned} &1+ \text{tan}^2 \ A = \text{sec}^2 \ A\\\\ &\text{tan}^2 \begin{pmatrix} \dfrac{\pi}{3} \end{pmatrix} - \text{sec}^2\begin{pmatrix} \dfrac{\pi}{3} \end{pmatrix} =-1 \end{aligned}\)