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:

sinθ=abcosθ=cbtanθ=ac \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:

sin1(ab)=θcos1(cb)=θtan1(ac)=θ \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 sin1 \sin^{-1} might be confusing, as we normally use a negative exponent to indicate the reciprocal. However, in this case, sin1α1sinα \sin^{-1} \alpha \neq \frac{1}{\sin \alpha} . When we want the reciprocal of sin \sin we use csc \csc . In order to avoid this ambiguity, sometimes people might choose to write the inverse functions with an arc prefix. For example:

arccosβ=cos1β \arccos \beta = \cos^{-1} \beta


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

cos1171130=tan1ab \cos^{-1} \frac{17}{\sqrt{1130}}= \tan^{-1} \frac{a}{b}

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

tan(cos1171130)=tan(tan1ab)tan(cos1171130θ)=ab \begin{aligned} \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{aligned}

Further, we we can use the ratio given to sketch the triangle with θ \theta in it, using the definition of cos1 \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 172+a2=1130 17^2 + a^2 = 1130 . This means a=1130289=29 a=\sqrt{1130-289}=29 . Finally, we evaluate tanθ=2917 \tan \theta = \frac{29}{17} , which means a+b=46 a+b=46 . _\square

Note by Arron Kau
7 years, 4 months ago

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Can you please give me inverse formulas of sine cosine tan

Anurag Suryawanshi - 7 years, 3 months ago

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