# More about Inverse Trigonometric Ratios

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In this section, let's discuss about the allowed inputs and possible outputs for inverse trigonomteric ratios.

## Allowed Inputs: Think: What will be the value of $${\sin ^{ - 1}}2$$? Is there any angle whose sine is 2?

The answer is NO! Since there is no angle whose sine is 2, it is absurd to talk about $${\sin ^{ - 1}}2$$. We can only apply the $${\sin ^{ - 1}}$$ operation to those numbers which can actually be generated by the sine function.

For example, the following expressions are meaningless:

\begin{align}&{\sin ^{ - 1}}\left( {\sqrt 3 } \right), \quad{\sin ^{ - 1}}\left( 7 \right)\\&{\sin ^{ - 1}}\left( { - \pi } \right),\quad{\sin ^{ - 1}}\left( { - 100} \right)\end{align}

The following expressions are all well-defined:

\begin{align}&{\sin ^{ - 1}}\left( {\frac{2}{{\sqrt 5 }}} \right),\quad {\sin ^{ - 1}}\left( {\frac{7}{{13}}} \right)\\&{\sin ^{ - 1}}\left( { - \frac{{17}}{{19}}} \right), \quad {\sin ^{ - 1}}\left( { - \sqrt {\frac{2}{3}} } \right)\end{align}

To summarize, the $${\sin ^{ - 1}}$$ operation can be applied to only those numbers which fall in the interval $$\left[ { - 1,1} \right]$$. The same holds true for the $${\cos ^{ - 1}}$$ operation. However, since the tan function generates output over the set of all real numbers, the $${\tan ^{ - 1}}$$ operation can be applied to any real number $$x$$, since there will always be an angle $$\theta$$ whose tan is $$x$$.

Using a similar line of reasoning, we conclude that:

• The $${\cot ^{ - 1}}$$ operation can be applied to any real number.

• The $${\sec ^{ - 1}}$$ and $${{\mathop{\rm cosec}\nolimits} ^{ - 1}}$$ operation can only be applied to numbers in the range $$\left( { - \infty , - 1} \right] \cup \left[ {1,\infty } \right)$$, since the $$\sec$$ and $${\text{cosec}}$$ functions produce outputs in this range only.

## Possible Outputs:

Let us analyze the characteristics of the outputs generated by the various inverse trigonometric ratio operations.

As an example, observe that

$\sin \frac{\pi }{6} = \frac{1}{2},\,\,\,\sin \frac{{5\pi }}{6} = \frac{1}{2}$

What value should we then assign to $${\sin ^{ - 1}}\left( {\frac{1}{2}} \right)$$? Isn’t it correct to say that $${\sin ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{6}$$ and also that $${\sin ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{{5\pi }}{6}$$? In fact, since there are an infinite number of angles whose sine is $$\frac{1}{2}$$, the $${\sin ^{ - 1}}$$ operation applied to $$\frac{1}{2}$$ should seemingly give infinitely many values.

Similar remarks exist for the other inverse trigonometric operations. For example,

• $${\cos ^{ - 1}}\left( 1 \right)$$ should give us all those angles whose cos is 1.

• $${\tan ^{ - 1}}\left( {\sqrt 3 } \right)$$ should give us all those angles whose tan is $$\sqrt 3$$.

• …and so on

However, for a number of reasons, we restrict the output of the inverse trigonometric operations in a way so that a unique output is generated in each case. For example, we will always write $${\sin ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{6}$$ and not $$\frac{{5\pi }}{6}$$.

How do we decide which value to pick in any such situation? There is a principal range for each inverse trigonometric operation, and the output must lie in that principal range.

The principal ranges of the various inverse trigonometric operations are shown in the table below:

 $${\sin ^{ - 1}}$$ $$\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$$ $${\cos ^{ - 1}}$$ $$\left[ {0,\pi } \right]$$ $${\tan ^{ - 1}}$$ $$\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$$ $${\cot ^{ - 1}}$$ $$\left( {0,\pi } \right)$$ $${\sec ^{ - 1}}$$ $$\left[ {0,\pi } \right] - \left\{ {\frac{\pi }{2}} \right\}$$ $${{\mathop{\rm cosec}\nolimits} ^{ - 1}}$$ $$\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right] - \left\{ 0 \right\}$$

Note: The reason behind selecting these particular intervals for the principal range will become clear at a later stage.

## Solved Examples:

Example 1: Which of the following are not well-defined mathematically? (a) $${\sin ^{ - 1}}\left( {\frac{1}{7}} \right)$$ (b) $${\cos ^{ - 1}}\left( {\sqrt 5 } \right)$$ (c) $${\tan ^{ - 1}}\left( { - \frac{1}{{1000}}} \right)$$ (d) $${\cot ^{ - 1}}\left( { - 99} \right)$$ (e) $${\sec ^{ - 1}}\left( {\frac{1}{3}} \right)$$ (f) $$\text{cosec}{^{ - 1}}\left( { - \sqrt {11} } \right)$$.

Solution: The terms in (a), (c), (d) and (f) are well-defined. The term in (b) is not well defined since the argument is greater than 1. The term in (e) is not well-defined since the magnitude of the argument is less than 1.

Example 2: Identify the incorrect values:

(a) $${\sin ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right) = \frac{{3\pi }}{4}$$

(b) $${\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) = \frac{\pi }{6}$$

(c) $${\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right) = \frac{{7\pi }}{6}$$

Solution: The values in (a) and (c) are incorrect, because they lie outside the principal range. The correct values should have been:

${\sin ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right) = \frac{\pi }{4},\,\,\,{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right) = \frac{\pi }{6}$ Challenge: Find the correct values of

(a) $${\sin ^{ - 1}}\left( { - \frac{1}{{\sqrt 2 }}} \right)$$

(b) $${\cos ^{ - 1}}\left( { - \frac{1}{2}} \right)$$

(c) $${\tan ^{ - 1}}\left( { - 1} \right)$$

Tip: The values should lie in the principal range of respective trigonomteric ratios.

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