# 3.11 Trigonometric functions  (Page 2/5)

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$f\left(x\right)=\mathrm{sec}\left(x\right)$

Again, the function is not defined for all real number “x”. Let us recall that :

$\mathrm{sec}x=\frac{1}{\mathrm{cos}\left(x\right)}$

This is a rational polynomial form, which is defined for $\mathrm{cos}\left(x\right)\ne 0$ . Now, cos(x) evaluates to zero for values of “x”, which appears at a certain interval given by the condition,

$\mathrm{cos}\left(x\right)=0;\phantom{\rule{1em}{0ex}}x=\left(2n+1\right)\frac{\pi }{2},\phantom{\rule{1em}{0ex}}\text{where}\phantom{\rule{1em}{0ex}}n\in Z$

The function cos(x) is zero for $x=±\frac{\pi }{2,}±\frac{3\pi }{2,}±\frac{5\pi }{2,}\dots$ etc. In other words, the cosine function is zero for all odd multiples of “ $\pi /2$ ”. It means that secant function is not defined for odd multiples of “ $\pi /2$ ”. These values of “x”, for which cosine is zero, need to be excluded from real number set “R”.

The values of secant function are bounded by certain intervals. We have seen that values of cosine function is between “-1” and “1”, including end points. Just like the case of cosecant function, the range of secant function is :

$|\mathrm{sec}x|\ge 1$

or

$\left(-\infty ,-1\right]\cup \left(1,\infty \right]$

Hence, domain and range of secant function are :

$\text{Domain}=R-\left\{x:x=\left(2n+1\right)\pi /2,\phantom{\rule{1em}{0ex}}n\in Z\right\}$

$\text{Range}=\left(-\infty ,-1\right]\cup \left(1,\infty \right]$

The plot of secant(x) .vs. x is shown here.

The period of secx is 2π. Important to note here is that function is not defined even within a periodic segment. Since sec(-x) = secx, we conclude that secant function is even function in each of periodic segment. Multiplication of secant function by a constant A changes the range plot lies on or beyond -A or A. The range is modified as :

$\text{Range}=\left(-\infty ,-A\right]\cup \left(A,\infty \right]$

Multiplying argument x like sec(kx), however, changes the points where function is not defined. It is now given by :

$x=\left(2n+1\right)\pi /2k,\phantom{\rule{1em}{0ex}}n\in Z$

Therefore, domain is now modified as :

$\text{Domain}=R-\left\{x:x=\left(2n+1\right)\pi /2k,\phantom{\rule{1em}{0ex}}n\in Z\right\}$

## Cotangent function

For a real number “x”, there is a cotangent function defined as :

$f\left(x\right)=\mathrm{cot}\left(x\right)$

The function is not defined for all real number “x”. Let us recall that :

$⇒\mathrm{cot}x=\frac{\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}$

This is a rational polynomial form, which is defined for $\mathrm{sin}\left(x\right)\ne 0$ . Now, sin(x) evaluates to values of “x”, which appears at a certain interval given by the condition,

$\mathrm{sin}\left(x\right)=0;\phantom{\rule{1em}{0ex}}x=n\pi ,\phantom{\rule{1em}{0ex}}\text{where}\phantom{\rule{1em}{0ex}}n\in Z$

This means that sin(x) is zero for $x=0,±\pi ,±2\pi ,±3\pi ,\dots$ etc. In other words, the sine function is zero for all integral multiples of “ $\pi$ ”. It means that cotangent function is not defined for integral multiples of “ $\pi$ ”. Values of “x”, for which sine is zero, need to be excluded from real number set “R”. On the other hand, the values of cotangent function are extended along the real number line on either side of zero. The range of the function, therefore, is “R”. Hence, domain and range of cotangent function are :

$\text{Domain}=R-\left\{x:x=n\pi ,\phantom{\rule{1em}{0ex}}n\in Z\right\}$

$\text{Range}=R$

The plot of cot(x) .vs. x is shown here.

The period of cotx is π. Since cot(-x) = -cotx, we conclude that cotangennt function is odd function in each of periodic segment. Multiplication of cotangennt function by a constant A does not change the range plot extends either side of x-axis. Multiplying argument x like cot(kx), however, changes the points where function is not defined. It is now given by :

$x=\frac{n\pi }{k},\phantom{\rule{1em}{0ex}}n\in Z$

Therefore, domain is now modified as :

$\text{Domain}=R-\left\{x:x=\frac{n\pi }{k},\phantom{\rule{1em}{0ex}}n\in Z\right\}$

## Examples

Problem : Find range of the function :

$f\left(x\right)=\frac{1}{2-\mathrm{sin}2x}$

Solution : The denominator of given function is non-negative as value of sin2x can not exceed 1. We know range of sin2x. We shall build up expression from this basic trigonometric function to determine range of the given function. Here,

$-1\le \mathrm{sin}2x\le 1$

Multiplying with -1 to change sign of sine function, we have (note the change in inequality sign) :

$1\ge -\mathrm{sin}2x\ge -1$ $2+1\ge 2-\mathrm{sin}2x\ge 2-1$ $3\ge 2-\mathrm{sin}2x\ge 1$

We need to take reciprocal of each term in the equality to obtain required function form (note the change in inequality sign),

$\frac{1}{3}\le \frac{1}{2-\mathrm{sin}2x}\le 1$ $\frac{1}{3}\le f\left(x\right)\le 1$ $\text{Range}=\left(\frac{1}{3,}1\right)$

This is a unique method to determine range by building up a function from a basic function along with change in the interval of values. We need to be careful that such building up of function does not introduce condition in which function becomes indeterminate. Further, we can find range conventionally by solving function for x in terms of y. We have not considered this method here as solution for x involves inverse trigonometric function. We shall, however, revisit this problem subsequent to the study of inverse trigonometric function.

Problem : Find range of function :

$f\left(x\right)=2\mathrm{sin}\sqrt{\left(\frac{{\pi }^{2}}{4}-{x}^{2}\right)}$

Solution : Before we attempt to find range, we need to find domain of the function so that we can determine interval of function values. We know that expression within square root is non-negative. Also, expression is a quadratic function. Analyzing this quadratic function, domain of quadratic function is found as [-π/2, π/2]. Coefficient of squared term is negative. Hence, its graph opens down and maximum value of quadratic function is :

${y}_{\mathrm{max}}=-\frac{D}{4a}=-\frac{0-\left(4X-1X\frac{\pi }{2}\right)}{4X-1}=\frac{\pi }{2}$

Since expression is non-negative within square root, minimum value of function is 0 (see figure). Now, sine function is an increasing function in [0, π/2] as is evident from its graph. Thus sine function assumes values in the interval [sin0, sin π/2]i.e. [0,1]. Sine function, however, has a coefficient of 2 . As such, range of given function is [0,2].

## Exercise

Problem : Find the domain of the function given by :

$f\left(x\right)=\sqrt{\mathrm{cos}\left(\mathrm{sin}x\right)}$

Solution : The argument (input) to cosine function is sine function. The expression within square root is non-negative. It means that :

$⇒\mathrm{cos}\left(\mathrm{sin}x\right)\ge 0$

We know that cosine function is positive in first and fourth quadrants. It means that the argument of the cosine function should be between -π/2 and π/2. Therefore, we need to see whether the value of “sinx” falls within this interval or not? The value of sine function, on the other hand, lies in the interval [-1,1]. This is indeed (as shown in the figure below) within the required interval for cosine function to be non-negative as 1<π/2 and -1>-π/2. The range of sine function falls within domain of cosine function

Now, we know that sine function is real for all real values of “x”. Hence, domain of the given function is :

$\text{Domain}=\left(-\infty ,\infty \right)$

Problem : Find the domain of the function given by :

$f\left(x\right)=\mathrm{cos}\frac{2\pi }{\left[x-1\right]}$

Solution : The cosine function is valid for all real values of its argument. The argument, however, is in rational form, requiring that denominator is not zero. Hence,

$\left[x-1\right]\ne 0$

We can easily evaluate this inequality knowing the fact that greatest integer function is zero in the interval 0≤x<1. This is also substantiated by the graph of greatest integer function as shown here. Now, applying to the greatest integer function of the denominator, the interval in which greatest integer function is equal to zero is : The greatest integer function [x] returns zero in the interval 0≤x<1.

$0\le x-1<1\phantom{\rule{1em}{0ex}}⇒1\le x<2$

Hence, domain of the given function is :

$\text{Domain}=R-\left[1,2\right)$

Check the validity of the composition and find domain of the function given by :

$f\left(x\right)=\mathrm{sin}\left[\mathrm{log}e\left\{\frac{\sqrt{\left(9-{x}^{2}\right)}}{1-x}\right\}\right]$

We know that range of logarithmic function is “R”. Here, logarithmic function itself is the argument of sine function. This means that argument of sine function is "R". This meets the requirement of a sine function, whose domain is "R". Hence, composition of function as given in the question is a valid composition.

In order to find, domain of the function, we need to find values of “x” for which argument of logarithmic function is a positive real number. In the nutshell, we need to evaluate following inequality :

$⇒\frac{\sqrt{\left(9-{x}^{2}\right)}}{1-x}>0$

Here, we see that numerator is a square root of a polynomial and is, therefore, positive. Now, evaluating of the polynomial in the numerator for positive real number, we have :

$⇒9-{x}^{2}>0$

$⇒{x}^{2}-9<0$

$⇒\left(x+3\right)\left(x-3\right)<0$

$⇒-3

Since numerator is positive, the denominator needs to be positive so that total rational polynomial is positive. Evaluating inequality relation for the polynomial in the denominator for positive real number :

$⇒1-x>0$

$⇒x<1$

Now, we know that the domain of the quotient is ${D}_{1}\cap {D}_{2}$ . Hence,

$\text{Domain}=\left(-3

$⇒\text{Domain of "f"}=-3

Find range of function :

$f\left(x\right)=\mathrm{cos}\sqrt{\left({x}^{2}-\frac{{\pi }^{2}}{9}\right)}$

Range is [-1,1]

Find range of function :

$f\left(x\right)=\mathrm{tan}\sqrt{\left(\frac{{\pi }^{2}}{9}-{x}^{2}\right)}$

Range is [0,√3]

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