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Draw the graph of y=|sinx| and hence the...

Draw the graph of `y=|sinx|` and hence the graph of `y=sin^(-1)|sinx|`.

Text Solution

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The graph of `y=sinx` is as follows.
Therefore, the graph of `y=|sinx|` can be drawn as shown in the following figure.

From the above graph, it is clear that `y=|sinx|` is non-differentiable for `x=npi,ninZ` Now `y=f(x)=sin^(-1)|sinx|`.
Now it is easy to draw the graph of `y=f(x)=sin^(-1)|sinx|` from the above graph.
`f(0)=0,f(pi//2)=pi//2,f(pi)=0`. Also the period of `y=f(x)` is `pi`.
Hence the graph of `y=f(x)` is as shown in the following figure.
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