Inverse Sine: Overview, Questions, Preparation

The inverse function of the sine (also called arcsin) is the inverse of the function of the sin. Since the sine function’s angle is equal to the opposite side ratio and the hypotenuse ratio, the angle measurement is thus given by the inverse sin of the same ratio.

The inverse or inverse equation function is used to calculate the angle calculation using the right triangle’s simple trigonometric ratios. Mostly, using sin-1, the inverse sine function is represented. This does not mean that the sin is not lifted to negative force.

A sin feature of an angle θ is equivalent to the opposite side of θ separated by hypotenuse in a right-angle triangle.

Sin θ = side/Hypotenuse

This is the underlying formula for the function of the sin.

Sin θ = On the opposite/Hypotenuse

We learnt about the sine function in the above explanation, where if the opposite side and hypotenuse are known to us, we can evaluate the sin from any angle.

The inverse sin or Sin-1 function takes the Opposite Side/Hypotenuse Side ratio and induces the angle. It is also written in the form of Arcsin.

The opposite sin is defined as sin-1 or arcsin.

Θ = Sin-1 (Hypotenuse/Opposite side)

Inverse Sine in Class 12:

In Class 12, this topic comes under the chapter “Inverse Trigonometric Functions,” which is a part of Unit-1 “Relations and Functions.” This unit carries 10 marks in the exams.

1. Solve arcsin (-√3/2).

Let y = arcsin(- √3 / 2). 

sin y = - √3 / 2 , with - π / 2 ≤ y ≤ π / 2

sin (π / 3) = √3 / 2.

We also know that sin(-x) = - sin x. 

So, sin (- π / 3) = - √3 / 2

Comparing this with the equation sin y = - √3 / 2, we can conclude that

y = - π / 3

2. Find sin−1 (sin 10)

Solution:

We know that sin−1 (sin θ) = θ, if - π2 ≤ θ ≤ π2.

Here, θ = 10 radians, which does not lie between - π2 and π2. 

But 3π - θ i.e., 3π - 10 lies between - π2 and π2 and sin (3π - 10) = sin 10.

Now, sin−1 (sin 10)

= sin-1 (sin (3π - 10)

= 3π - 10

Therefore, sin−1 (sin 10) = 3π - 10.

3. Find sin (cos−1 3/5)

Solution:  

Let, cos−1 3/5 = θ 

Therefore, cos θ = 3/5

Therefore, sin θ = √(1 - cos2  θ) = √(1 - 9/25) = √(16/25) = 4/5.

Therefore, sin (cos−1 3/5) = sin θ = 4/5.

Q: What is sine inverse?

Q: What is the inverse-sin value of 1?

Q: What is Arcsin?

Q: Is the inverse sine equal to the function of Cosec?

Q: Why is the inverse sin confined?