# Possible Outputs For Inverse T-Ratios

Possible Outputs For Inverse T-Ratios

Let us analyze the characteristics of the outputs generated by the various inverse T-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 trig 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 trig 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 trig operation, and the output must lie in that principal range.

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

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

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

Example 1: 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:

${\sin ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right) = \frac{\pi }{4},\,\,\,{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right) = \frac{\pi }{6}$

Example 2: Find the correct values of

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

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

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

Solution: We have:

\begin{align}&{\sin ^{ - 1}}\left( { - \frac{1}{{\sqrt 2 }}} \right) = - \frac{\pi }{4}\\&{\cos ^{ - 1}}\left( { - \frac{1}{2}} \right) = \pi - \frac{\pi }{3} = \frac{{2\pi }}{3}\\&{\tan ^{ - 1}}\left( { - 1} \right) = - \frac{\pi }{4}\end{align}

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More Important Topics
Numbers
Algebra
Geometry
Measurement
Money
Data
Trigonometry
Calculus
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