# Discover the Solution Set for sin^-1 x > cos^-1 x with Greatest Integer Function

• utkarshakash
In summary, the solution set for [sin^-1 x]>[cos^-1 x] where [] denotes greatest integer function is (sin1,1). This is found by taking into account that sin^(-1)(x) = a and cos^(-1)(x) = b are angles, with -π/2 <= a <= π/2 and 0 <= b <= π. The square bracket "rounds down", so [a] = greatest integer <= a. Thus, [a] and [b] must be integers lying in the ranges -π/2 < [a] < π/2 and 0 <= [b] < π (with [b] >= 0 rather than [b] >
utkarshakash
Gold Member

## Homework Statement

Find the solution set of [sin^-1 x]>[cos^-1 x] where [] denotes greatest integer function.

## The Attempt at a Solution

I know that
-∏/2 <sin^-1 x < ∏/2
0 < cos^-1 x <∏

But I am clueless what will happen if I enclose them within "those square brackets"!

Those square brackets look to me like a havoc when it comes to these types of questions.
These types of questions are best done by sketching the graph of the functions.

utkarshakash said:

## Homework Statement

Find the solution set of [sin^-1 x]>[cos^-1 x] where [] denotes greatest integer function.

## The Attempt at a Solution

I know that
-∏/2 <sin^-1 x < ∏/2
0 < cos^-1 x <∏

But I am clueless what will happen if I enclose them within "those square brackets"!

Remember that sin^(-1)(x) = a and cos^(-1)(x) = b are angles, with -π/2 <= a <= π/2 and 0 <= b <= π. The square bracket "rounds down", so [a] = greatest integer <= a. Thus, [a] and must be integers lying in the ranges -π/2 < [a] < π/2 and 0 <= < π (with >= 0 rather than > 0). What are all the integers lying in these two ranges? Now, by checking a few cases you can find the appropriate ranges of x.

Ray Vickson said:
Remember that sin^(-1)(x) = a and cos^(-1)(x) = b are angles, with -π/2 <= a <= π/2 and 0 <= b <= π. The square bracket "rounds down", so [a] = greatest integer <= a. Thus, [a] and must be integers lying in the ranges -π/2 < [a] < π/2 and 0 <= < π (with >= 0 rather than > 0). What are all the integers lying in these two ranges? Now, by checking a few cases you can find the appropriate ranges of x.

For a the integral values can be -1,0,1 and for b it can be 0,1,2,3. Now I can see that a will be greater than b only if a=1 and b=0. So the range of x comes out be (sin1,1). Thanks!

## 1. What is the solution set for sin^-1 x > cos^-1 x with Greatest Integer Function?

The solution set for this inequality is all real numbers between -1 and 1, excluding the values where sin^-1 x and cos^-1 x are equal. This can be represented as: (-∞, sin^-1 x) U (cos^-1 x, ∞).

## 2. How do I graph this inequality?

To graph this inequality, first graph the equations y = sin^-1 x and y = cos^-1 x separately. Then, use the Greatest Integer Function to round down the values of both functions. Finally, shade the area where the rounded down value of sin^-1 x is greater than the rounded down value of cos^-1 x.

## 3. Can this inequality have more than one solution?

Yes, this inequality can have multiple solutions. Since both sin^-1 x and cos^-1 x have a range of -π/2 to π/2, there can be an infinite number of values between these ranges that satisfy the inequality.

## 4. What is the domain of this inequality?

The domain of this inequality is all real numbers between -1 and 1. This is because both sin^-1 x and cos^-1 x have a domain of -1 to 1.

## 5. How can I solve this inequality analytically?

To solve this inequality analytically, you can use the properties of trigonometric functions and the Greatest Integer Function. First, rewrite the inequality as sin^-1 x - cos^-1 x > 0. Then, use the inverse trigonometric identities and the properties of the Greatest Integer Function to simplify the expression and find the solution set.

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