3.11 Trigonometric functions  (Page 5/5)

$$f\left(x\right)=\sec \left(x\right)$$

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

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

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

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

The function cos(x) is zero for $x=\pm \frac{\pi }{\mathrm{2,}}\pm \frac{3\pi }{\mathrm{2,}}\pm \frac{5\pi }{\mathrm{2,}}\dots$ 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 :

$$\left|\sec x\right|\ge 1$$

or

$$\left(-\infty ,-1\right]\cup \left(\mathrm{1,}\infty \right]$$

Hence, domain and range of secant function are :

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

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

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

Secant function

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 :

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

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

$$x=\left(2n+1\right)\pi /\mathrm{2k},n\in Z$$

Therefore, domain is now modified as :

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

Cotangent function

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

$$f\left(x\right)=\cot \left(x\right)$$

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

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

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

$$\sin \left(x\right)=0;x=n\pi ,\text{where}n\in Z$$

This means that sin(x) is zero for $x=\mathrm{0,}\pm \pi ,\pm 2\pi ,\pm 3\pi ,\dots$ 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 :

$$\text{Domain}=R-\{x:x=n\pi ,n\in Z\}$$

$$\text{Range}=R$$

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

Cotangent function

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 :

$$x=\frac{n\pi }{k},n\in Z$$

Therefore, domain is now modified as :

$$\text{Domain}=R-\{x:x=\frac{n\pi }{k},n\in Z\}$$

Examples

Exercise

Problem : Find the domain of the function given by :

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

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

$$\Rightarrow \cos \left(\sin x\right)\ge 0$$

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.

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 :

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

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,

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

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 :

Greatest integer function

$$0\le x-1<1\Rightarrow 1\le x<2$$

Hence, domain of the given function is :

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