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Inverse Trigonometric Functions Formulas

The inverse trigonometric functions are called as arcus functions or anti trigonometric functions. These are the inverse functions of the trigonometric functions with domains. Here, we will study the Inverse Trigonometric Functions Formulas for the sine, cosine, tangent, cotangent, secant, and the cosecant functions, and are used to obtain an angle from any of the angle’s trigonometric ratios.

Inverse Trigonometric Function Formulas

What is Inverse Trigonometric Function?

The inverse trigonometric functions are also known as the anti trigonometric functions or arcus functions. The inverse trigonometric functions of sine, cosine, tangent, cosecant, secant, and cotangent are used to find the angle of a triangle from any of the trigonometric functions. It is widely used in many fields like geometry, engineering, physics, etc.

But in most of the time, the convention symbol to represent the inverse trigonometric function using arc-prefix like $$\arcsin(x)$$, $$\arccos(x)$$, $$\arctan(x)$$ etc. The inverse trigonometric functions are written as $$\sin^{-1}x,$$ $$\cos^{-1} x$$, $$\cot^{-1} x$$, $$\tan^{-1} x$$, $$\csc^{-1} x$$, $$\sec^{-1} x$$. Now let us get the formulas related to these functions. Consider, the function y = f(x), and x = g(y) then the inverse function is written as g = f-1.

Source: en.wikipedia.org

Important Inverse Trigonometric Functions Formulas

• $$\sin^{-1} (-x) = – \sin^{-1} (x), x ∈ [-1, 1]$$
• $$\cos^{-1} (-x) = \pi – \cos^{-1} (x), x ∈ [-1, 1]$$
• $$\tan^{-1} (-x) = – \tan^{-1} (x), x ∈ R$$
• $$\csc^{-1} (-x) = -\csc^{-1} (x), |x| ≥ 1$$
• $$\sec^{-1} (-x) = \pi -\sec^{-1} (x), |x| ≥ 1$$
• $$\cot^{-1} (-x) = \pi – \cot^{-1} (x), x ∈ R$$
• $$\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} , x ∈ [-1, 1]$$
• $$\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}, x ∈ R$$
• $$\sec^{-1}x + \csc^{-1} x = \frac{\pi}{2},|x| ≥ 1$$
• $$\sin^{-1} (1/x) = \csc^{-1} (x), if x ≥ 1 or x ≤ -1$$
• $$\cos^{-1} (1/x) = \sec^{-1} (x), if x ≥ 1 or x ≤ -1$$
• $$\tan^{-1} (1/x) = \cot^{-1} (x), x > 0$$
• $$\tan^{-1} x + \tan^{-1} y = \tan^{-1} (\frac{(x+y)}{(1-xy)})$$
• $$\tan^{-1} x – \tan^{-1} y = \tan^{-1}(\frac{ (x-y)}{(1+xy)})$$
• $$\sin(\sin^{-1} (x)) = x, -1≤ x ≤1$$
• $$\cos(\cos^{-1} (x)) = x, -1≤ x ≤1$$
• $$\tan(\tan^{-1} (x)) = x, – ∞ < x < ∞$$
• $$\csc(\csc^{-1} (x)) = x, – ∞ < x ≤ 1 or -1 ≤ x < ∞$$
• $$\sec(\sec^{-1} (x)) = x,- ∞ < x ≤ 1 or 1 ≤ x < ∞$$
• $$\cot(\cot^{-1} (x)) = x, – ∞ < x < ∞$$
• $$\sin^{-1} (\sin θ) = θ, -\frac{\pi}{2}≤ θ ≤ \frac{\pi}{2}$$
• $$\cos^{-1} (\cos θ) = θ, 0 ≤ θ ≤ \pi$$
• $$\tan^{-1}(\tan θ) = θ, -\frac{\pi}{2}< θ < \frac{\pi}{2}$$
• $$\csc^{-1} (\csc θ) = θ, – \frac{\pi}{2} ≤ θ < 0 or 0 < θ ≤ \frac{\pi}{2}$$
• $$\sec^{-1} (\sec θ) = θ, 0 ≤ θ ≤ \frac{\pi}{2}or \frac{\pi}{2}< θ ≤ \pi$$
• $$\cot^{-1} (\cot θ) = θ, 0 < θ < \pi$$

Solved Examples for Inverse Trigonometric Functions Formulas

Q1. Find the values of $$\tan^{-1} \sin (-\frac{\pi}{2})$$

Solution:

$$\tan^{-1} \sin (-\frac{\pi}{2})$$

$$= \tan^{-1} (- \sin \frac{\pi}{2})$$

$$= \tan^{-1} (- 1), [Since – \sin \frac{\pi}{2}= -1]$$

$$= \tan^{-1} (- \tan \pi/4), [Since tan \frac{\pi}{4}= 1]$$

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

$$= – \frac{\pi}{4}.$$

Therefore, $$\tan^{-1} \sin (-\frac{\pi}{2}) = – \frac{\pi}{4}$$

Q2. Find the values of $$\sin (\cos^{-1}\frac{3}{5})$$

Solution:

Let, $$\cos^{-1}\frac{3}{5} = \Theta$$

$$\cos \Theta = \frac{3}{5}$$

$$\sin \Theta = \sqrt {(1 – cos2 \Theta)}$$

$$= \sqrt { (1 – \frac{9}{25}) }$$

$$= \sqrt {\frac{16}{25} }$$

$$= \frac{4}{5}$$

Therefore, $$\sin (\cos^{-1}\frac{3}{5})$$

$$= \sin \Theta = \frac{4}{5}.$$

Q3. Find the values of $$\cos (\tan^{-1}\frac{3}{4})$$

Solution:

Let, $$\tan^{-1}\frac{3}{4}= \Theta$$

$$\tan \Theta = \frac{3}{4}$$

We know that $$sec2 \Theta – tan2 \Theta = 1$$

$$\sec \Theta = \sqrt {(1 + tan2 \Theta)}$$

$$\sec \Theta = \sqrt {(1 + (\frac{3}{4})2)}$$

$$\sec \Theta = \sqrt {(1 + \frac{9}{16})}$$

$$\sec \Theta = \sqrt { (\frac{25}{16})}$$

$$\sec \Theta = \frac{5}{4}$$

$$\ cos \Theta = \frac{4}{5}$$

$$\Theta = \cos^{-1}\frac{4}{5}$$

Now, $$\cos (\tan^{-1}\frac{3}{4}) = \cos (\cos^{-1}\frac{4}{5}) = \frac{4}{5}$$

Therefore, $$\cos (\tan^{-1}\frac{3}{4}) = \frac{4}{5}$$

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