Simplify the expression $\sqrt{(3-\pi)^2}$. Then, consider the condition $\sin(3-\pi) < 0$.

AlgebraAbsolute ValueSimplificationTrigonometryInequalitiesSquare Roots
2025/3/11

1. Problem Description

Simplify the expression (3π)2\sqrt{(3-\pi)^2}. Then, consider the condition sin(3π)<0\sin(3-\pi) < 0.

2. Solution Steps

We are asked to simplify (3π)2\sqrt{(3-\pi)^2}. We know that x2=x\sqrt{x^2} = |x|. Therefore,
(3π)2=3π \sqrt{(3-\pi)^2} = |3-\pi|
Since π3.14159\pi \approx 3.14159, we know that 3π<03-\pi < 0. Thus, 3π=(3π)=π3|3-\pi| = -(3-\pi) = \pi-3.
Also, we are given that sin(3π)<0\sin(3-\pi) < 0. Since 3π3 - \pi is a negative number and π3.14\pi \approx 3.14, 3π0.143-\pi \approx -0.14. Since sin(x)\sin(x) is negative when xx is negative, this inequality is satisfied.
Thus, (3π)2=3π=π3\sqrt{(3-\pi)^2} = |3-\pi| = \pi-3.

3. Final Answer

π3\pi - 3