```sqrt (a^2-x^2)+a*sin^-1(x/a)= ?`

`` ``

a larger than zero

Expert Answers

An illustration of the letter 'A' in a speech bubbles

Let us assume that `sin^-1 (x/a) = y`

==> `siny = x/a`

`` But we know that sin(y) = opposite/hypotenuse = `x/a`

Then, we will assume that we have a right angle triangle such that x is the opposite side and a is the hypotenuse.

Then, the adjacent side = `sqrt(a^2 - x^2)`

`` ==> Adjacent = `sqrt(a^2-x^2)`

`` ==> `siny= x/a`

`` ==> `cosy = sqrt(a^2-x^2)/a`

`` ==> `sqrt(a^2-x^2)= acosy = acos^-1 (x/a).......(1)`

`` ==> Now we will rewrite the equation

==> `sqrt(a^2-x^2) + asin^-1 (x/a)`  .

==> `acos^2 -1(x/a) + asin^-1 (x/a)`

`` ==> `a (cos^-1 (x/a) + sin^ -1 (x/a))`

`` ==> But we know that `sin^-1 a + cos^-1 a = pi/2`

`` ==> `sqrt(a^2+x^2) + sin^-1 (x/a) = a*pi/2`

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial