# 3.14 Inverse trigonometric functions  (Page 4/4)

 Page 4 / 4

## Summary

Redefined domains of trigonometric functions are tabulated here :

--------------------------------------------------------------------------------------------------- Trigonometric Old New Old NewFunction Domain Domain Range Range ---------------------------------------------------------------------------------------------------sine R [-π/2, π/2] [-1,1][-1,1] cosine R [0, π][-1,1] [-1,1]tan R – odd multiples of π/2 (-π/2, π/2) R R cosecant R – integer multiple of π [-π/2, π/2]– {0} R – (-1,1) R – (-1,1) secant R - odd multiples of π/2 [0, π]– {π/2} R – (-1,1) R – (-1,1) cotangent R – integer multiple of π (0, π) R R---------------------------------------------------------------------------------------------------

We observe that there is no change in the range – even though domains of the trigonometric functions have changed.

The corresponding domain and range of six inverse trigonometric functions are tabulated here.

-------------------------------------------------------------- Inverse Domain RangeTrigonometric Function-------------------------------------------------------------- arcsine [-1,1][-π/2, π/2]arccosine [-1,1] [0, π]arctangent R (-π/2, π/2)arccosecant R – (-1,1) [-π/2, π/2] – {0}arcsecant R – (-1,1) [0, π] – {π/2}arccotangent R (0, π) --------------------------------------------------------------

## Exercise

Find the domain of the function given by :

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

The exponent of the exponential function is inverse trigonometric function. Exponential function is real for all real values of exponent. We see here that given function is real for the values of “x” corresponding to which arcsine function is real. Now, domain of arcsine function is [-1,1]. This is the interval of "x" for which arcsine is real. Hence, domain of the given function, “f(x)” is :

$\text{Domain}=\left[-1,1\right]$

Problem : Find the domain of the function given by :

$f\left(x\right)={\mathrm{cos}}^{-1}\frac{3}{3+\mathrm{sin}x}$

Solution : The given function is an inverse cosine function whose argument is a rational function involving trigonometric function. The domain interval of inverse cosine function is [-1, 1]. Hence, value of argument to inverse cosine function should lie within this interval. It means that :

$-1\le \frac{3}{3+\mathrm{sin}x}\le 1$

Comparing with the form of modulus, $|x|\le 1\phantom{\rule{1em}{0ex}}⇒-1\le x\le 1$ , we conclude :

$⇒\frac{|3|}{|3+\mathrm{sin}x|}\le 1$

Since, modulus is a non-negative number, the inequality sign remains same after simplification :

$⇒|3|\le |3+\mathrm{sin}x|$

Again 3>0 and 3+sin x>0, we can open up the expression within the modulus operator without any change in inequality sign :

$⇒3\le 3+\mathrm{sin}x\phantom{\rule{1em}{0ex}}⇒0\le \mathrm{sin}x\phantom{\rule{1em}{0ex}}⇒\mathrm{sin}x\ge 0$

The solution of sine function is the domain of the given function :

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

Find range of the function :

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

We have already solved this problem by building up interval in earlier module. Here, we shall find domain conventionally by solving for x. The denominator of given function is non-negative as value of sin2x can not exceed 1. Hence, domain of function is real number set R. Further, maximum value of sin2x is 1. Hence,

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

This means given function is positive for all real x. Now, solving for x,

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

We know that domain of sine inverse function is [-1,1]. Hence,

$-1\le \frac{2y-1}{y}\le 1$

Since y>0, we can simplify this inequality as :

$-y\le 2y-1\le y$

Either,

$⇒2y-1\ge -y$ $⇒y\ge \frac{1}{3}$

Or,

$⇒2y-1\le y$ $⇒y\le 1$

$\text{Range}=\left[\frac{1}{3},1\right]$

Find domain of function

$f\left(x\right)={\mathrm{sin}}^{-1}\left\{{\mathrm{log}}_{2}\left({x}^{2}+3x+4\right)\right\}$

This is a composite function in which quadratic function is argument of logarithmic function. The logarithmic function is, in turn, argument of inverse sine function. In such case, it is advantageous to evaluate from outer to inner part. The domain of outermost inverse trigonometric function is [-1,1].

$-1\le \left\{{\mathrm{log}}_{2}\left({x}^{2}+3x+4\right)\right\}\le 1$ ${\mathrm{log}}_{2}{2}^{-1}\le \left\{{\mathrm{log}}_{2}\left({x}^{2}+3x+4\right)\right\}\le {\mathrm{log}}_{2}2$ $\frac{1}{2}\le \left({x}^{2}+3x+4\right)\le 2$

For the first inequality,

$⇒2{x}^{2}+6x+8\ge 1$ $⇒2{x}^{2}+6x+7\ge 0$

This quadratic function is positive for all value of x. For the second inequality,

$⇒{x}^{2}+3x+4\le 2$ $⇒{x}^{2}+3x+2\le 0$

The solution of this inequality is [1,2]. The intersection of R and [1,2]is [1,2]. Hence, domain of given function is [1,2].

Find the range of the function

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

Hint : The range of rational expression as argument of inverse trigonometric function is [0,1]. But, domain of arccosine is [-1,1]and range is [0,π]. The function is continuously decreasing. The maximum and minimum values are 0 and1 (see arccosine graph). Hence, range of given function is [0, π/2].

Find the range of the function

$f\left(x\right)={\mathrm{cosec}}^{-1}\left[1+{\mathrm{sin}}^{2}x\right]$

where [.] denotes greatest integer function.

The minimum and maximum value of ${\mathrm{sin}}^{2}x$ is 0 and 1. Hence, range of $1+\mathrm{sin}{}^{2}x$ is defined in the interval given by :

$1\le 1+{\mathrm{sin}}^{2}x\le 2$

The corresponding values returned by GIF are 1 and 2. It means :

$\left[1+{\mathrm{sin}}^{2}x\right]=\left\{1,2\right\}$

But domain of arccosecant is [-π/2, π/2] – {0}. Refer graph of arccosecant. Thus, arccosecant can take only 1 as its argument, which falls within the domain of arccosecant. Hence, range of given function is a singleton :

$\text{Range}=\left\{{\mathrm{cosec}}^{-1}1\right\}$

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