which function represents y = sin x ?
which function represents y = sin x ?
Answer
Explanation:
Step1: Recall the properties of ( y = \sin x )
The sine function ( y=\sin x ) has a period of ( 2\pi ), passes through the origin ((0,0)), and at ( x = \frac{\pi}{2} ), ( y = 1 ); at ( x=\pi ), ( y = 0 ); at ( x=\frac{3\pi}{2} ), ( y=-1 ); at ( x = 2\pi ), ( y = 0 ). Also, the graph of ( y=\sin x ) is an odd function, symmetric about the origin, and it crosses the ( x )-axis at ( x = n\pi ) for integer ( n ), and has a "wave" shape starting from the origin, going up to ( ( \frac{\pi}{2},1) ), down to ( (\frac{3\pi}{2}, - 1) ), etc.
Step2: Analyze each graph
- First graph: It seems to be symmetric about the ( y )-axis (even function), which is the graph of ( y = \cos x ) (since ( \cos x ) is even), so not ( \sin x ).
- Second graph: Does not pass through the origin (at ( x = 0 ), ( y\neq0 )), so not ( \sin x ).
- Third graph: Let's check the origin. At ( x = 0 ), the graph passes through ( (0,0) )? Wait, no, looking at the third graph, when ( x = 0 ), the value is not 0? Wait, no, re - check. Wait, the fourth graph: Wait, let's re - evaluate. Wait, the standard ( y=\sin x ) at ( x = 0 ) is 0. Let's check the fourth graph: At ( x = 0 ), the graph passes through ( (0,0) )? Wait, no, the third and fourth: Wait, the standard ( y = \sin x ) has the following behavior: when ( x=0 ), ( y = 0 ); when ( x=\frac{\pi}{2} ), ( y = 1 ); when ( x=\pi ), ( y = 0 ); when ( x=\frac{3\pi}{2} ), ( y=-1 ). Let's check the fourth graph: At ( x = 0 ), the graph is at 0? Wait, no, the fourth graph: Let's see the direction. The sine function starts at the origin, goes up to ( \frac{\pi}{2} ), then down. The fourth graph: When ( x = 0 ), it's at 0? Wait, no, the third graph: Wait, maybe I mixed up. Wait, the standard ( y=\sin x ) has the graph that, for ( x>0 ), first increases from 0 to ( \frac{\pi}{2} ) (to ( y = 1 )), then decreases to 0 at ( \pi ), then decreases to - 1 at ( \frac{3\pi}{2} ), then increases to 0 at ( 2\pi ). The fourth graph: Let's check the key points. At ( x=\frac{\pi}{2} ), does it reach 1? At ( x = \frac{\pi}{2} ), the fourth graph seems to have a peak at ( x=\frac{\pi}{2} ) (y = 1), at ( x=\pi ), y = 0, at ( x=\frac{3\pi}{2} ), y=-1, and passes through the origin. Wait, the third graph: At ( x = 0 ), the graph is at 0? Wait, no, the third graph: When ( x = 0 ), the value is 0? Wait, maybe I made a mistake. Wait, the standard ( y=\sin x ) is an odd function, so it should satisfy ( f(-x)=-f(x) ). Let's check the fourth graph: For ( x=-\frac{\pi}{2} ), ( y=-1 ) (since ( \sin(-\frac{\pi}{2})=-1 )), and for ( x=\frac{\pi}{2} ), ( y = 1 ), which satisfies ( f(-x)=-f(x) ). The third graph: At ( x=-\frac{\pi}{2} ), ( y = 1 ), and at ( x=\frac{\pi}{2} ), ( y=-1 ), which is also odd, but let's check the direction. The standard ( y=\sin x ) for ( x>0 ) starts by increasing from 0. The fourth graph: When ( x ) is a small positive number, the function value is positive (since it goes up to ( \frac{\pi}{2} )), while the third graph, for small positive ( x ), goes negative? Wait, no. Wait, the standard ( y=\sin x ) at ( x = \frac{\pi}{4} ) (a small positive ( x )) has ( y=\frac{\sqrt{2}}{2}>0 ). So the graph that, for ( x>0 ), starts by increasing from 0 (going up) is the correct one. The fourth graph: At ( x = 0 ), ( y = 0 ), for ( x=\frac{\pi}{2} ), ( y = 1 ), for ( x=\pi ), ( y = 0 ), for ( x=\frac{3\pi}{2} ), ( y=-1 ), which matches the properties of ( y=\sin x ). Wait, maybe I messed up the third and fourth. Wait, let's re - check the third graph: At ( x = 0 ), the graph is at 0? Wait, the third graph: When ( x = 0 ), the point is (0,0)? Wait, no, the third graph's curve at ( x = 0 ) seems to be at 0? Wait, maybe the fourth graph is the one. Wait, the standard ( y=\sin x ) has the graph that, when ( x ) increases from 0, ( y ) increases to 1 at ( \frac{\pi}{2} ), then decreases to 0 at ( \pi ), then decreases to - 1 at ( \frac{3\pi}{2} ), then increases to 0 at ( 2\pi ). The fourth graph: Let's see the direction. From ( x=- \frac{3\pi}{2} ) to ( x = -\pi ), it's increasing, from ( x=-\pi ) to ( x = -\frac{\pi}{2} ), decreasing to - 1, then from ( x=-\frac{\pi}{2} ) to ( x = 0 ), increasing to 0, then from ( x = 0 ) to ( x=\frac{\pi}{2} ), increasing to 1, then decreasing to 0 at ( \pi ), then decreasing to - 1 at ( \frac{3\pi}{2} ), which matches the sine function's behavior. The third graph: From ( x = 0 ) to ( x=\frac{\pi}{2} ), it's decreasing, which is the opposite of ( \sin x ) (which should increase from 0 to ( \frac{\pi}{2} )). So the fourth graph (the bottom - right graph) is the graph of ( y=\sin x ).