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

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svjr | Student, Undergraduate | (Level 2) Honors

Posted October 26, 2011 at 7:32 AM via web

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

`` ``

a larger than zero

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted October 26, 2011 at 8:07 AM (Answer #1)

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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`

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