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.

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  • Trigonometric Ratios of Standard Angles

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Main Content

  • 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:

polar

 

{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.

 

triangle

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}

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