The sine function, sin(x), has a range of [−1,1]. This means that the value of sin(θ) will always be between −1 and 1, inclusive, for any real number θ.
The inequality you provided is:
0≤sin(
43748131634
x
2
+y
2
)
This inequality is asking for the values of x and y for which the sine of the expression is greater than or equal to zero.
Let θ=
43748131634
x
2
+y
2
.
The inequality becomes 0≤sin(θ), which is true when θ is in the intervals [2nπ,(2n+1)π] for any integer n.
Since x
2
+y
2
≥0, the value of θ is also always non-negative.
Therefore, we only need to consider non-negative intervals for θ.
So, 2nπ≤
43748131634
x
2
+y
2
≤(2n+1)π for n=0,1,2,…
This inequality is true for a specific set of x and y values, not for all possible x and y values. For example, if we choose x and y such that the value of the sine function is negative, the inequality will not hold.
For example, let
43748131634
x
2
+y
2
2
3π
.
Then sin(
2
3π
)=−1, and the inequality 0≤−1 is false.
Since the inequality is not true for all possible values of x and y, it’s not a universal truth.
Based on the nature of the inequality, none of the provided options (A, B, or C) seem to apply, as they appear to be related to a different context. The inequality is a mathematical statement that can be true or false depending on the values of the variables. Therefore, D. 以上都不是 (None of the above) is the most appropriate answer.