Inverse Sine Function

The inverse sine function is one of the inverse trigonometric functions which determines the inverse of the sine function and is denoted as sin^-1 or Arcsine.  For example: If the value of sine 90 degree is 1, then the value of inverse sin 1 or sin^-1(1) will be equal to 90°. Each trigonometric function such as cosine, tangent, cosecant, cotangent has its inverse in a restricted domain. The inverse function formulas are used to calculate the measurement of angles with the help of the trigonometric ratios from the right-angle triangle. Generally, the sine inverse function is represented as sin^-1. It does not state that sine cannot be raised to the negative power.

You can easily derive the sine inverse function formula. Students can calculate the inverse of trigonometric function through the calculators also. You can easily find out the inverse of the function if you are aware of the six basic trigonometric functions. Through this article, you will study what inverse sine, inverse sine function, inverse sine graph, inverse sine derivation, inverse sine examples, inverse sine formulas etc. are.

Inverse Sine Function

To understand the concept of inverse sine function clearly, we should first know about sine function.

Sine Function - The sine function of angle ϴ in the right-angle triangle is defined as the ratio of the opposite side of angle ϴ to the hypotenuse side.

Sin\[\theta = \frac{\text{Opposite side}}{\text{Hypotenuse side}}\]

The inverse of sin function or sin^-1 , also known as arcsin or asine obtains the angle ϴ if it takes the ratio \[\frac{\text{Opposite side}}{\text{Hypotenuse side}}\]

Sine Inverse is represented by  sin^-1 or arcsin.

Sin Inverse Example

In a triangle PQR, PQ = 4.9 cm, QR = 4.0 cm and PR = 2.8 CM and angle Q = 35°

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Solution: Sin 35° = \[\frac{\text{Opposite side}}{\text{Hypotenuse side}}\]

Sin 35° = \[\frac{2.8}{4.9}\]

Sin 35° = 0.57

So, sin^-1 \[\frac{\text{Opposite side}}{\text{Hypotenuse side}}\] = 35°

Sin^-1 (0.57)= 35°

Inverse Sine Formula

Let us understand the inverse sine formula through an example: 

Let us consider that you want to calculate the depth of the deep sea-bed from the bottom of the sea, and the following two specifications are given:

• The angle through which the cable makes with the deep sea–bed.

• The length of the cable

The sine function helps you to calculate the distance or depth of the ship from the deep sea-bed through the following method. 

If the cable’s length is given as 40 m and the angle is 39°, then

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Sin 39° = \[\frac{\text{Opposite side}}{\text{Hypotenuse side}}\]

Sin 39° = \[\frac{d}{40}\]

Depth (d) = Sin 39° x 40

Depth (d) = 0.6293 x 40

Depth (d) = 25.172

Hence, the depth of the sea-bed is 25.17 cm.

Trigonometric Sine Function and Sine Inverse Formula Summary

The formula for trigonometric sine function for angle ϴ is stated as

Sin\[\theta = \frac{\text{Opposite side}}{\text{Hypotenuse side}}\]

The inverse sine formula is stated as

Sin\[^{-1}\] = \[\frac{\text{Opposite side}}{\text{Hypotenuse side}}\] = \[\theta\] 

Inverse Sine Graph

Arcsine function, also known as the inverse of the sine function, is represented as sin⁻¹ x. It is represented in the graph as shown below:(image will be uploaded soon)

Inverse Sine Derivative

Let us represent the function by ϴ = sin⁻¹ k. We will make use of implicit differentiation to compute the derivative of this function.

Eliminate the inverse from the function

ϴ = sin⁻¹ k → sin ϴ = k

The new function will be written in the form of k = sin ϴ

Where the domain of ϴ is restricted to the range of the principal values that the sin⁻¹ function can select.

Differentiate the equation on both sides in terms of ϴ

\[\frac{dk}{d\theta }=\frac{d}{\theta}\left ( sin\theta  \right )\]

\[\frac{dk}{d\theta }=\left ( cos \theta  \right )\]

Through the trigonometric identity, sin²ϴ + cos²ϴ = 1 we get,

\[\frac{dk}{d\theta }=\sqrt{1-sin^{2}\theta }\]

Using the inverse function, we can write the equations as,

\[\frac{dk}{d\theta }=\sqrt{1-k^{2} }\]

Through the reciprocals of derivatives, we get

\[\frac{dk}{d\theta }=\frac{1}{\frac{dk}{d\theta }}\]

Here, you can see the values of inverse sine in tabulated form.

Inverse Sine Value Table

Derivative of Sin Inverse x w.r.t. Cos Inverse √(1- x²)

There is another way of checking the derivative of sin inverse x. We have now calculated the derivative of sin inverse x to be 1/√(1- x²), where -1 < x < 1. Now, we will proceed to differentiate sin⁻¹ x, concerning another function. This function is cos-1√(1-x²). To continue our calculations, we will assume that cos-1√(1- x²) is equal to some variable, such as z. 

Assume y = sin-1x

This implies that sin y = x

Using cos² θ + sin² θ = 1, we now have cos θ = √(1 - sin² θ)

This means that cos y = √(1 - sin² y) = √(1- x²)

Upon differentiating sin y = x with respect to x, we will get

cos y (dy/dx) = 1

⇒ dy/dx = 1/cos y

⇒ dy/dx = 1/√(1- x²). Let this be equation 1.

Now, assume z = cos-1√(1- x²) ⇒ sin z = x and cos z = √(1- x²) 

Now, differentiating cos z = √(1- x²) w.r.t. x, we have -sin z (dz/dx) = -2x/2√(1- x²)

⇒ -x(dz/dx) = -x/√(1- x²)

⇒ dz/dx = 1/√(1- x²)

⇒ dx/dz = √(1- x²). Let this be equation 2. 

Now, to be able to determine the value of d(sin⁻¹ x)/d(cos⁻¹ √(1- x²)) = dy/dz,

We see that dy/dz = dy/dx × dx/dz = [1/√(1- x²)] × √(1 - x²) = 1

Thus, the derivative of sin inverse x with respect to cos inverse √(1 - x²) is 1.

Proof of Derivative of Sin Inverse x

We are now aware that the derivative of sin inverse x is 1/√(1 - x²), where x lies between -1 and 1. This derivative of sin inverse x can easily be derived using some differentiation formulae. Let us see exactly how we can reach this derivation. 

We know, cos² θ + sin² θ= 1,

The chain rule states that (f(g(x)))' = f'(g(x)).g'(x)

Thus, d(sin x)/dx = cos x

Assume y = sin⁻¹ x ⇒ sin y = x

Differentiating both sides of sin y = x with respect to x, we get

cos y dy/dx = 1

This implies that dy/dx = 1/cos y

⇒ dy/dx = 1/√(1 - sin² y) (Using cos² θ + sin² θ is equal to 1)

⇒ dy/dx = 1/√(1 - x²) (Because sin y is equal to x)

⇒ d(sin⁻¹ x)/dx = 1/√(1 - x²)

Hence, d(sin⁻¹ x)/dx = 1/√(1 - x²). We have proved that the derivative of sin inverse x is 1/√(1 - x²).

Anti-Derivative of Sin Inverse x

There is another topic that we are yet to deal with. This is calculating the antiderivative of sin inverse x. It will not be wrong to state that the anti-derivative of sin inverse x is nothing but the integral of sin inverse x. We know that to find the antiderivative of any given value, we have to follow the inverse process of differentiation. After a series of calculations, it has been found that the anti-derivative of sin inverse x is x sin⁻¹ x + √(1-x²) + C, where C is referred to as the constant of integration. In this manner, we have been successful in obtaining the antiderivative of sin inverse x and sin x + C.

Thus, to sum it up, ∫ sin⁻¹ x = x sin⁻¹ x + √(1- x²) + C