# Trigonometric Functions

Preliminaries

• We were introduced to the trigonometric functions in class ${XI}$ with their general definitions for any angle ${\theta}$. The specialty of these functions is that their values get repeated after an interval. This is because after every interval of ${2 \pi^c}$, the angles have same initial and terminal arms.
• Basics : If we fix the initial arm of the angle ${\theta}$ as positive ${X}$ axis, with vertex at origin. Let ${P (x,y)}$ be ANY point on the terminal arm of the angle. Let ${r}$ be ${\sqrt {x^2 + y^2}}$. Then,

${sin (\theta) = \frac {y}{r}, \ cos (\theta) = \frac {x}{r}, \ tan (\theta) = \frac {y}{x}, \ cosec (\theta) = \frac {r}{y}, \ sec (\theta) = \frac {r}{x}, \ cot (\theta) = \frac {x}{y}}$

• Periodicity : The characteristic of trigonometric functions, which repeat their values after a fixed interval, is known as periodicity. The smallest non-negative interval is known as the fundamental period.

• Trigonometric Ratios of Standard Angles

• Equations and Solutions

The value which satisfies an equation is known as a solution. For example, ${2x+4=0}$ is an equation. On simplifying,

${2x= -4,}$

${So, \ x = \frac {-4}{2} = -2}$

So, the solution is ${x= -2}$.

• Trigonometric Equations

The equations involving trigonometric functions are known as trigonometric equations. The unknown values, which are to be found, represent the measure of an angle. For example, ${sin (\theta)= \frac 1 {\sqrt 2}}$ is satisfied by ${\theta = \frac {\pi}{4}}$ or ${45^o}$.

• Principal Solutions

The solution, which lies between ${0}$ and ${2 \pi}$, is known as the principal solution.

${0 \le \theta < 2 \pi}$

• General Solutions

We saw that the measure of full circle is ${2 \pi}$ radians or ${360^o}$. Hence, after adding ${2 \pi}$ to above angle, we will still get a solution. So,

${sin (2 \pi + \frac {\pi}{4}) = \frac {1}{\sqrt 2}}$

Similarly,

${sin (4 \pi + \frac {\pi}{4}) = \frac {1}{\sqrt 2}}$

Thus, there are infinitely many solutions to above equations, apart from ${\theta = \frac \pi 4}$. These solutions are known as general solutions. Consider another example:

${sin \ \theta = 0 \Rightarrow \theta = n \pi, n \in \mathbb Z}$

Note: We will have to use allied angles formulas and the formula sheet to solve problems of this kind, where we are asked to find the general solutions.

• Polar Coordinates

The Cartesian coordinates ${(x,y)}$ and the polar coordinates ${(r, \theta)}$ are inter-convertible. See the figure below:

${x= r \ cos \ \theta, y = r \ sin \ \theta }$

${r^2 = x^2 + y^2, tan \ \theta = \frac y x, \theta = tan^{-1} \frac y x}$

• Solving a Triangle

A triangle has 3 sides, say, ${a,b,c}$ and 3 angles, ${\angle A, \angle B, \angle C}$. By solving a triangle, we mean, to find values of all lengths of sides and all angles of a triangle. See the figure below.

The side opposite to ${\angle A}$ is denoted by ${a}$ and so on.

There are certain rules to be used to solve these problems. These are:

• Sine Rule

${\dfrac {a}{sin \ A} = \dfrac {b}{sin \ B}= \dfrac {c}{sin \ C}=2 R}$

${R}$ is the radius of the circumcircle.

• Cosine Rule

${a^2 = b^2 + c^2 -2bc cos \ A}$

${b^2 = a^2 + c^2 -2ac cos \ B}$

${c^2 = a^2 + b^2 - 2ab cos \ A}$

• Projection Rule

${a = b cos \ C + c cos \ B}$

${b = a cos \ C + c cos \ A}$

${c = a cos \ B + b cos \ A}$

• Half Angle Formulas

${a+b+c = 2s, \ s}$ is the semi-perimeter.

${sin \ \dfrac A 2 = \sqrt {\dfrac {(s-b)(s-c)}{bc}}}$

${cos \ \dfrac A 2 = \sqrt {\dfrac {s(s-a)}{bc}}}$

${tan \ \dfrac A 2 = \sqrt { \dfrac {(s-b)(s-c)}{s(s-a)}}}$

• Area of a Triangle

The area of triangle is given by,

${area = \dfrac 1 2 \ bc \ sin \ A = \dfrac 1 2 \ ac \ sin \ B = \dfrac 1 2 \ ab \ sin \ C}$

It is also given by

${area = \sqrt {s(s-a)(s-b)(s-c)}}$. This is known as Heron’s formula.

• Napier’s Analogies

${tan \left ( \dfrac {B-C}{2} \right ) = \dfrac {b-c}{b+c}cot \ \dfrac A 2}$

• Inverse Trigonometric Functions

If a function ${f}$ is defined from set A to set B (which is one-one and onto), we can define an inverse function, ${f^{-1}}$ from set B to set A. So, if ${x \in A, y \in B}$, then ${y = f(x)}$ and ${x = f^{-1} y}$.

If ${sin \ \dfrac {\pi}{6} = \dfrac 1 2}$, then ${sin^{-1} \frac 1 2 = \dfrac \pi 6}$. If ${cos \ \dfrac {\pi}{4} = \dfrac 1 {\sqrt 2}}$, then ${sin^{-1} \dfrac 1 {\sqrt 2} = \dfrac \pi 4}$.

Note that ${sin^{-1} y}$ is different from ${[sin (y)]^{-1}}$. ${[sin (y)]^{-1}}$ is ${\frac 1 {sin (y)}}$, which is ${cosec (y)}$.

On the other hand, ${sin^{-1} y}$ is the angle, whose sine is ${y}$.

Corresponding to each of six trigonometric functions, we have 6 inverse trigonometric functions, i.e. ${sin^{-1} y, cos^{-1}y, tan^{-1}y, sec^{-1}y, cosec^{-1}y, cot^{-1}y}$.

Recall : Principal value is the value of angle, which lies between ${0}$ and ${2 \pi}$.

• Properties of Inverse Trigonometric Functions

Set I

${sin^{-1} \dfrac 1 x = cosec^{-1}x, \ for \ x \ge 1, x \le -1}$

${cos^{-1} \dfrac 1 x = sec^{-1}x, \ for \ x \ge 1, x \le -1}$

${tan^{-1} \dfrac 1 x = cot^{-1} x, for \ x > 0}$

Set II

${sin^{-1} (-x) = - sin^{-1}x, -1 \le x \le 1}$

${tan^{-1} (-x) = - tan^{-1}x, x \in \mathbb R}$

${cosec^{-1} (-x) = - cosec^{-1} x, for \ x \ge 1, x \le -1}$

Set III

${cos^{-1} (-x) = \pi - cos^{-1}x, -1 \le x \le 1}$

${cot^{-1} (-x) = \pi - cot^{-1}x, x \in \mathbb R}$

${sec^{-1} (-x) =\pi - sec^{-1} x, for \ x \ge 1, x \le -1}$

Set IV

${sin^{-1}(x)+ cos^{-1}(x) = \dfrac {\pi}{2}, -1 \le x \le 1}$

${tan^{-1}(x)+ cot^{-1}(x) = \dfrac {\pi}{2}, x \in \mathbb R}$

${cosec^{-1}(x)+ sec^{-1}(x) = \dfrac {\pi}{2}, -1 \le x \le 1}$

Set V

${tan^{-1}x + tan^{-1}y = tan^{-1} \dfrac {x+y}{1-xy}, when \ x,y > 0 \ and \ xy<1}$

${tan^{-1}x + tan^{-1}y = \pi + tan^{-1} \dfrac {x+y}{1-xy}, when \ x,y > 0 \ and \ xy > 1}$

${tan^{-1}x - tan^{-1}y = tan^{-1} \dfrac {x-y}{1+xy}, when \ x,y > 0 }$

Set VI

${2 tan^{-1}x = sin^{-1} \dfrac {2x}{1+x^2}, for \ -1 \le x \le 1}$

${2 tan^{-1}x = cos^{-1} \dfrac {1-x^2}{1+x^2}, for \ x \ge 0}$

${2 tan^{-1}x = tan^{-1} \dfrac {2x}{1-x^2}, for \ -1 < x < 1}$