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

[utkarshakash]

Homework Statement

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

Homework Equations

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"!

 

[Saitama]

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.

 

[Ray Vickson]

Homework Statement

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

Homework Equations

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.

 

[utkarshakash]

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!

PS- Can you please help me out on my other problems?