# Inverse Trigonometric Ratios

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## Introduction:

Suppose that we apply the sine function to a certain angle $$\theta$$, and obtain the output as $$y$$. We can write

$y = \sin \theta$

This equation expresses $$y$$ in terms of $$\theta$$. Can we invert this relation to writing $$\theta$$ in terms of $$y$$?

Let us consider an analogous algebraic example. Consider the equation

$y = 3x + 2$

This equation expresses $$y$$ as a function of $$x$$. It is easy to invert this and express $$x$$ as a function of $$y$$:

\begin{align}3x &= y - 2\\ \Rightarrow x &= \frac{{y - 2}}{3} \end{align} Think: Can we do something similar when we have trigonometric ratios?

## Inverse Trigonometry:

The answer to is yes – we have the notations of inverse trigonometry to help us. For example, we can invert the relation by writing

$\theta = {\sin ^{ - 1}}y$

Note: Do not confuse $${\sin ^{ - 1}}y$$ with $$\frac{1}{{\sin y}}$$. The latter is the reciprocal of $$\sin y$$. The former is completely different: the term  $${\sin ^{ - 1}}y$$  is to be interpreted as an angle value whose sine will be $$y$$.

The following table lists some examples of the $${\sin ^{ - 1}}$$ operation:

 $$\sin 0 = 0$$ $${\sin ^{ - 1}}0 = 0$$ $$\sin \frac{\pi }{6} = \frac{1}{2}$$ $${\sin ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{6}$$ $$\sin \frac{\pi }{4} = \frac{1}{{\sqrt 2 }}$$ $${\sin ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right) = \frac{\pi }{4}$$ $$\sin \frac{\pi }{3} = \frac{{\sqrt 3 }}{2}$$ $${\sin ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) = \frac{\pi }{3}$$ $$\sin \frac{\pi }{2} = 1$$ $${\sin ^{ - 1}}1 = \frac{\pi }{2}$$

Here are some examples of the $${\cos ^{ - 1}}$$ operation:

 $$\cos 0 = 1$$ $${\cos ^{ - 1}}1 = 0$$ $$cos\frac{\pi }{6} = \frac{{\sqrt 3 }}{2}$$ $$co{s^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) = \frac{\pi }{6}$$ $$\cos \frac{\pi }{4} = \frac{1}{{\sqrt 2 }}$$ $$co{s^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right) = \frac{\pi }{4}$$ $$\cos \frac{\pi }{3} = \frac{1}{2}$$ $${\cos ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{3}$$ $$\cos \frac{\pi }{2} = 0$$ $${\cos ^{ - 1}}0 = \frac{\pi }{2}$$

And a few examples of the $${\tan ^{ - 1}}$$ operation

 $$\tan 0 = 0$$ $${\tan ^{ - 1}}0 = 0$$ $$\tan \frac{\pi }{6} = \frac{1}{{\sqrt 3 }}$$ $${\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right) = \frac{\pi }{6}$$ $$\tan \frac{\pi }{4} = 1$$ $${\tan ^{ - 1}}\left( 1 \right) = \frac{\pi }{4}$$ $$\tan \frac{\pi }{3} = \sqrt 3$$ $${\tan ^{ - 1}}\left( {\sqrt 3 } \right) = \frac{\pi }{3}$$

## Solved Examples:

Example 1: Find the values of

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

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

(c) $${\tan ^{ - 1}}\left( {\sqrt 3 } \right)$$

(d) $${\cot ^{ - 1}}\left( 1 \right)$$

(e) $${\sec ^{ - 1}}\left( 2 \right)$$

(f) $${{\mathop{\rm cosec}\nolimits} ^{ - 1}}\left( {\frac{2}{{\sqrt 3 }}} \right)$$

Solution:

(a) Since \begin{align}\sin \frac{\pi }{6} = \frac{1}{2}\end{align}, we have \begin{align}{\sin ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{6}\end{align}.

(b) Since \begin{align}\cos \frac{\pi }{4} = \frac{1}{{\sqrt 2 }}\end{align}, we have \begin{align}{\cos ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right) = \frac{\pi }{4}\end{align}.

(c) We know that \begin{align}\tan \frac{\pi }{3} = \sqrt 3 \end{align}, and so \begin{align}{\tan ^{ - 1}}\left( {\sqrt 3 } \right) = \frac{\pi }{3}\end{align}.

(d) Since \begin{align}\cot \frac{\pi }{4} = 1\end{align}, we have \begin{align}{\cot ^{ - 1}}\left( 1 \right) = \frac{\pi }{4}\end{align}.

(e) Since \begin{align}\sec 0 = 1\end{align}, we have \begin{align}{\sec ^{ - 1}}\left( 1 \right) = 0\end{align}.

(f) Since \begin{align}{\mathop{\rm cosec}\nolimits} \frac{\pi }{3} = \frac{2}{{\sqrt 3 }}\end{align}, we have \begin{align}{{\mathop{\rm cosec}\nolimits} ^{ - 1}}\left( {\frac{2}{{\sqrt 3 }}} \right) = \frac{\pi }{3}\end{align}.

Example 2: Find the value of $${\sin ^{ - 1}}\left( {\frac{1}{2}} \right) + {\cos ^{ - 1}}\left( {\frac{1}{2}} \right)$$.

Solution: We have:

\begin{align}&{\sin ^{ - 1}}\left( {\frac{1}{2}} \right) + {\cos ^{ - 1}}\left( {\frac{1}{2}} \right)\\&= \frac{\pi }{6} + \frac{\pi }{3} \\&=\boxed{ \frac{\pi }{2}}\end{align}

Example 3: Find the value of the expression $${\sin ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) + {\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right)$$ .

Solution: We have:

\begin{align}&{\sin ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) + {\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right)\\ &= \frac{\pi }{3} + \frac{\pi }{6} \\&= \boxed{\frac{\pi }{2}}\end{align}

The previous two examples have the same answer. Let’s explore this connection in the next problem.

Example 4: Simplify $${\sin ^{ - 1}}x + {\cos ^{ - 1}}x$$.

Solution: Suppose that $${\sin ^{ - 1}}x = \theta$$. This means that $$\sin \theta = x$$. Now, we can also write this equality as follows:

\begin{align}x &= \sin \theta = \cos \left( {\frac{\pi }{2} - \theta } \right)\\ \Rightarrow {\cos ^{ - 1}}x &= \frac{\pi }{2} - \theta \end{align}

Thus,

$\boxed{{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \theta + \left( {\frac{\pi }{2} - \theta } \right) = \frac{\pi }{2}}$ Challenge: Simplify the following:

• $${\tan ^{ - 1}}x + {\cot ^{ - 1}}x$$
• $${\sec ^{ - 1}}x + {\text{cosec} ^{ - 1}}x$$

Tip: Use a similar approach as in example-4.

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Grade 10 | Questions Set 1
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