Inverse Trigonometric Functions

Inverse Trigonometric Functions


Let’s keep it simple and straight. The quantities such as \(\sin{}^{-1}\alpha\), \(\cos{}^{-1}\alpha\), \(\tan{}^{-1}\alpha\), … are Inverse Trigonometric Functions. We also call them Inverse Circular or Cyclometric Functions. These are widely used inter alia in engineering and navigation.

You may choose to start with: Measurement of Angles

So, \(\sin{}^{-1}\alpha\) is an angle. Precisely, it denotes the smallest angle whose sine is \(\alpha\). That is, if \(\sin\theta = \alpha\), then \(\sin{}^{-1}\alpha\) is equal to the smallest angle \((\theta)\) that satisfies the relation. The angle may be positive or negative.

We know that for \(\sin\theta = \alpha\) where \(\alpha\) is a known definite quantity, \((\theta)\) is not a known definite angle but one of a series of angles.

To note,

  • \(\sin{}^{-1}\alpha\), \(\tan{}^{-1}\alpha\), \(\csc{}^{-1}\alpha\) and \(\cot{}^{-1}\alpha\) always lie between \(-90{°}\) and \(+90{°}\);
  • when \(\alpha\) is \(+\)ve, \(\sin{}^{-1}\alpha\) lies between \(0{°}\) and \(+90{°}\);
  • when \(\alpha\) is \(-\)ve, \(\sin{}^{-1}\alpha\) lies between \(-90{°}\) and \(0{°}\); and
  • \(\cos{}^{-1}\alpha\) and \(\sec{}^{-1}\alpha\) invariably lie between \(0{°}\) and \(180{°}\).

It should be clear by now that \(\sin{}^{-1}\alpha\) is not equivalent to \(\frac{1}{\sin(\alpha)}\). In fact, if we wish, we may write the later as \((\sin\alpha)^{-1}\). Moreover, we also write \(\sin{}^{-1}\alpha\), \(\cos{}^{-1}\alpha\), \(\tan{}^{-1}\alpha\), … as \(\arcsin(x)\), \(\arccos(x)\), \(\arctan(x)\), …

SL Loney – Examples 43

Prove these:

1. \(\sin{}^{-1}\frac{3}{5} + \sin{}^{-1}\frac{8}{17} = \sin{}^{-1}\frac{77}{85}\)

3. \(\cos{}^{-1}\frac{4}{5} + \tan{}^{-1}\frac{3}{5} = \tan {}^{-1}\frac{27}{11} \)

7. \(\tan {}^{-1}\frac{1}{2} + \tan {}^{-1}\frac{1}{3} = 45{°}\)

8. \(\tan {}^{-1}\frac{1}{7} + \tan {}^{-1}\frac{1}{13} = \tan {}^{-1}\frac{2}{9}\)

9. \(2\tan{}^{-1}\frac{2}{3} = \tan{}^{-1}\frac{12}{5}\)

13. \(\tan{}^{-1}\frac{1}{3} + \tan{}^{-1}\frac{1}{5} + \tan{}^{-1}\frac{1}{7} + \tan{}^{-1}\frac{1}{8} = \frac{\pi}{4}\)

17. \(\tan{}^{-1}\frac{m}{n} – \tan{}^{-1}\frac{m – n}{m + n} = \frac{\pi}{4}\)

39. Solve: \(\sin {}^{-1}x + \sin {}^{-1}2x = \frac{\pi}{3}\)

Solutions

#1.
Let \(\sin{}^{-1}\frac{3}{5} = A\)
\(\Rightarrow \sin A = \frac{3}{5}\) and \(\cos A = \frac{4}{5}\)

Similarly, for 2nd term:
\(\sin B = \frac{8}{17}\) and \(\cos B = \frac{15}{17}\)

\(\therefore \sin(A + B) = \sin A \cos B + \cos A \sin B = \frac{77}{85}\)

\(\Rightarrow A + B = \sin{}^{-1}\frac{77}{85}\)

Hence, \(\sin{}^{-1}\frac{3}{5} + \sin{}^{-1}\frac{8}{17} = \sin{}^{-1}\frac{77}{85}\)

#3.
Consider LHS:
Let 1st term \( = \cos \alpha\) and 2nd term \( = \tan \beta\)
then, \(\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 – tan\alpha \cdot tan\beta} = \frac{27}{11}\)

or, \(\alpha + \beta = \tan {}^{-1}\frac{27}{11}\)
Hence, \(\cos{}^{-1}\frac{4}{5} + \tan{}^{-1}\frac{3}{5} = \tan {}^{-1}\frac{27}{11} \)

#7.
Let \(\tan {}^{-1}\frac{1}{2} = \alpha\) and \(\tan {}^{-1}\frac{1}{3} = \beta\)
then, \(\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 – tan\alpha \cdot tan\beta} = 1 = \tan 45{°}\)

Hence, \(\tan {}^{-1}\frac{1}{2} + \tan {}^{-1}\frac{1}{3} = 45{°}\)

#8.
Let \[
\tan {}^{-1}\frac{1}{7} = \alpha \Rightarrow \tan \alpha = \frac{1}{7}
\] \[
\tan {}^{-1}\frac{1}{13} = \beta \Rightarrow \tan \beta = \frac{1}{13}
\] \[
\tan {}^{-1}\frac{2}{9} = \gamma \Rightarrow \tan \gamma = \frac{2}{9}
\] Now, \[
\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 – \tan(\alpha)\tan(\beta)} = \frac{\frac{1}{7} + \frac{1}{13}}{1 – (\frac{1}{7})(\frac{1}{13})} = \frac{2}{9}
\] \[
\Rightarrow (\alpha + \beta) = \tan {}^{-1}\frac{2}{9}
\] \[
\Rightarrow \tan {}^{-1}\frac{1}{7} + \tan {}^{-1}\frac{1}{13} = \tan {}^{-1}\frac{2}{9} = RHS
\]

Useful Link

#9.
Let \(\tan{}^{-1}\frac{2}{3} = \alpha \Rightarrow \tan(\alpha) = \frac{2}{3}\)

Now, \(\tan(2\alpha) = \frac{2\tan(\alpha)}{1 – \tan(\alpha)^2} = \frac{12}{9}\)
\(\therefore 2\alpha = \tan{}^{-1}\frac{12}{9}\)

Or, \(2\tan{}^{-1}\frac{2}{3} = \tan{}^{-1}\frac{12}{5}\)

#13.
Here, LHS \(= \tan{}^{-1}\frac{\frac{1}{3} + \frac{1}{5}}{1 – \frac{1}{3} \cdot \frac{1}{5}} + \tan{}^{-1}\frac{\frac{1}{7} + \frac{1}{8}}{1 – \frac{1}{7} \cdot \frac{1}{8}} \)

\(= \tan{}^{-1}\frac{4}{7} + \tan{}^{-1}\frac{3}{11} = \tan{}^{-1}\frac{\frac{4}{7} + \frac{3}{11}}{1 – \frac{4}{7} \cdot \frac{3}{11}} \)

\(= \tan{}^{-1}1 = \frac{\pi}{4} =\) RHS

#17.
LHS\(= \tan{}^{-1}\frac{\frac{m}{n} – \frac{m – n}{m + n}}{1 + \frac{m}{n} \cdot \frac{m – n}{m + n}} = \tan{}^{-1}1 = \frac{\pi}{4} =\) RHS

#39.
Let \[
\sin{}^{-1}x = \alpha \] then, \[ \sin\alpha = x
\] and \[
\cos\alpha = \sqrt{1 – x^2}
\]Similarly, from 2nd term \[
\sin\beta = 2x
\] and \[
\cos\beta = \sqrt{1 – 4x^2}
\]so, the eqn. becomes\[
\alpha + \beta = \frac{\pi}{3}
\]or, \[
\cos(\alpha + \beta) = \cos\frac{\pi}{3}
\]\[
\Rightarrow \cos\alpha\cos\beta – \sin\alpha\sin\beta = \frac{1}{2}
\]Putting values,\[
x = \pm \frac{1}{2}\sqrt{\frac{3}{7}}
\]

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