# How calculate f(2/5)+f(9/5)> given function domain [0,2] given function range [-90,90]f=sin^-1(1-x)

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You need to find values of function at x = `2/5 ` and x = `9/5` .

Plugging x = `2/5` in f(x) yields: `f(2/5)=sin^-1(1 - 2/5) =gt f(2/5)=sin^-1(3/5)`

Plugging `x = 9/5` in f(x) yields:`f(9/5)=sin^-1(1 - 9/5) =gt f(9/5)=sin^-1(-4/5) = pi - sin^-1(4/5)`

`f(2/5)+f(9/5) = sin^-1(3/5) + sin^-1(-4/5)`

`f(2/5)+f(9/5) ~~ 36.86^o + 180^o - 53.13^o`

`f(2/5)+f(9/5) ~~ 163^o`

**Hence, evaluating the value of the sum yields `f(2/5)+f(9/5) ~~ 163^o` .**