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Inverse Trigonometric Functions


Inverse Trigonometric Functions are, like any other inverse function, mathematical operators that undo another function's operation.

Given a triangle like this

Triangle ABC

Triangle ABC

the basic trigonometric functions would be defined as:

\[ \begin{array}{lcr} \sin \theta = \frac{a}{b} & \cos \theta = \frac{c}{b} & \tan \theta = \frac{a}{c} \end{array} \]

with the angle as their input (or argument) and a ratio of sides as their result. However, the inverse functions take the ratio as input and return the angle:

\[ \sin^{-1} \left( \frac{a}{b} \right) = \theta \\ \cos^{-1} \left( \frac{c}{b} \right) = \theta \\ \tan^{-1} \left( \frac{a}{c} \right) = \theta \]

This means the inverse trigonometric functions are useful whenever we know the sides of a triangle and want to find its angles.

Note: The notation \( \sin^{-1} \) might be confusing, as we normally use a negative exponent to indicate the reciprocal. However, in this case, \( \sin^{-1} \alpha \neq \frac{1}{\sin \alpha} \). When we want the reciprocal of \( \sin \) we use \( \csc \). In order to avoid this ambiguity, sometimes people might choose to write the inverse functions with an arc prefix. For example:

\[ \arccos \beta = \cos^{-1} \beta \]


In following statement, \( a \) and \( b \) are positive, co-prime integers. What is the sum of \( a \) and \( b \)?

\[ \cos^{-1} \frac{17}{\sqrt{1130}}= \tan^{-1} \frac{a}{b} \]

Since we are trying to find \( a \) and \( b \), we should take the tangent of both sides of the equation:

\[ \begin{align} \tan \left( \cos^{-1} \frac{17}{\sqrt{1130}} \right) &= \tan \left( \tan^{-1} \frac{a}{b} \right) \\ \tan \left( \underbrace{\cos^{-1} \frac{17}{\sqrt{1130}}}_{\large{\theta}} \right) &= \frac{a}{b} \end{align} \]

Further, we we can use the ratio given to sketch the triangle with \( \theta \) in it, using the definition of \( \cos^{-1} \):

Right Triangle with side 17 and hypotenuse square root 1130

Right Triangle with side 17 and hypotenuse square root 1130

Now, using the Pythagorean theorem, we can see that \( 17^2 + a^2 = 1130 \). This means \( a=\sqrt{1130-289}=29 \). Finally, we evaluate \( \tan \theta = \frac{29}{17} \), which means \( a+b=46 \). \( _\square \)

Note by Arron Kau
3 years, 5 months ago

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Can you please give me inverse formulas of sine cosine tan Anurag Suryawanshi · 3 years, 4 months ago

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