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 \(y\):

\[\begin{align}3x &= y - 2\\ \Rightarrow x &= \frac{{y - 2}}{3} \end{align}\]

enlightenedThink: Can we do something similar when we have trigonometric ratios?


Inverse Trigonometry:

The answer to enlightened 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,

\[\boxed{{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \theta  + \left( {\frac{\pi }{2} - \theta } \right) = \frac{\pi }{2}}\]


yesChallenge: 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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