Find the exact value of sin−1sin10π9.Write your answer in radians in terms of π.If necessary, click on "Undefined."

The function sin(x) is periodic with period 2π. This means that sin(x) = sin(x + 2πn) for any integer n.

The value 10π/9 is greater than π/2 but less than π, which means it falls in the second quadrant. In the second quadrant, the sine function is positive but the cosine function is negative.

However, the inverse sine function, sin^(-1)(x), is only defined in the range -π/2 to π/2. This means we need to find an equivalent angle for 10π/9 that falls within this range.

We can do this by subtracting π from 10π/9, which gives us -8π/9. This angle is in the range of the inverse sine function, and sin(-8π/9) = sin(10π/9) because of the periodicity of the sine function.

Therefore, sin^(-1)(sin(10π/9)) = -8π/9.