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10. 方程 arcsinx+arccosy=nπ(nZ)\arcsin x+\arccos y=n \pi \quad(n \in \mathbf{Z}) 所表示的图解是 ( ). (A) (B...
Feb 26, 2024
10. 方程 arcsinx+arccosy=nπ(nZ)\arcsin x+\arccos y=n \pi \quad(n \in \mathbf{Z}) 所表示的图解是 ( ). (A) (B) (C) (D)
Solution by Steps
step 2
The sum of the two functions arcsinx+arccosy\arcsin x + \arccos y will be within the range [π2,3π2][-\frac{\pi}{2}, \frac{3\pi}{2}]
step 3
Since nZn \in \mathbf{Z}, and the equation is arcsinx+arccosy=nπ\arcsin x + \arccos y = n\pi, the only possible values for nn that fall within the range are n=0n = 0 or n=1n = 1
step 4
For n=0n = 0, the equation becomes arcsinx+arccosy=0\arcsin x + \arccos y = 0. This is only possible if x=0x = 0 and y=1y = 1, which is a single point on the graph
step 5
For n=1n = 1, the equation becomes arcsinx+arccosy=π\arcsin x + \arccos y = \pi. This is true for all xx and yy such that x2+y2=1x^2 + y^2 = 1, which is the equation of a circle with radius 1
C
Key Concept
Sum of arcsin\arcsin and arccos\arccos
Explanation
The sum arcsinx+arccosy=nπ\arcsin x + \arccos y = n\pi represents specific geometric shapes depending on the value of nn. For n=0n = 0, it's a point, and for n=1n = 1, it's a circle with radius 1.
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